Given equation of plane is x + 2y + 3z = 2
Length of perpendicular from (4, 3, 2) to the plane is
\(d=\cfrac { 4+2\left( 3 \right) +3\left( 2 \right) }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } } } =\cfrac { 4+6+6 }{ \sqrt { 14 } } \)
= \(\cfrac { 14 }{ \sqrt { 14 } } =\cfrac { \sqrt { 14 } .\sqrt { 14 } }{ \sqrt { 14 } } =\sqrt { 14 } \) units
Let us find the image of the point (4,3,2) to the plane x + 2y + 3z = 2
Here \(\vec { u } =4\hat { i } +3\hat { j } +2\hat { k } ,\vec { n } =\hat { i } +2\hat { j } +3\hat { k } \)
Then the image \(\vec { v } =\vec { u } +\cfrac { 2\left[ p-\left( \vec { u } .\vec { n } \right) \right] }{ \left| \vec { n } \right| ^{ 2 } } \)
\(\vec { v } =\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) +\cfrac { 2\left[ 2-\left( 4+6+6 \right) \right] }{ \left( \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } } \right) } \left( \hat { i } +2\hat { j } +3\hat { k } \right) \)
= \(\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) +\frac { 2\left( 2-16 \right) }{ 14 } \left( \hat { i } +2\hat { j } +3\hat { k } \right) \)
= \(\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) +\cfrac { 2\left( -14 \right) }{ 14 } \left( \hat { i } +2\hat { j } +3\hat { k } \right) \)
= \(\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) -2\left( \hat { i } +2\hat { j } +3\hat { k } \right) \)
= \(\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) -2\left( \hat { i } +4\hat { j } -6\hat { k } \right) \)
= \(2\hat { i } -\hat { j } -4\hat { k } \)
\(\therefore\) The foot of the \(\bot \) from (4, 3, 2) to the plane is
\(\cfrac { \left( 4\hat { i } +3\hat { j } +2\hat { k } \right) +\left( 2\hat { i } -\hat { j } -4\hat { k } \right) }{ 2 } \)
= \(\cfrac { 6\hat { i } +2\hat { j } -2\hat { k } }{ 2 } =3\hat { i } +\hat { j } -\hat { k } \)
Hence, the co-ordinates of the foot of the perpendicular is (3, 1, -1)

Given equation of line is
\(\frac { x-1 }{ 1 } =\frac { y }{ 2 } =z+1=s\)
\(\Rightarrow x-1=s\Rightarrow x=s+1\)
\(\frac { y }{ 2 } =s\Rightarrow y=2s\)
\(z+1=s\Rightarrow z=s-1\)
\(\therefore\)  Any point on the line is of the form (s+1, 2s, s-1)
This point lies on the plane 2x-y+2z = 2
\(\Rightarrow 2(s+1)-(2s)+2(s-1)=2\)

\(\\ \Rightarrow 2s=2\Rightarrow s=1\)
when s = 1, the point is (1 + 1, 2(1), 1 - 1)
= (2, 2, 0)
\(sin\ \theta = \frac{|\vec b.\vec n|}{|\vec b||\vec n|} \)
\(\vec b = \vec i+2\vec j+\vec k, \quad \vec n = 2\vec i-\vec j+ 2\vec k \)
\(\vec b.\vec n = 2-2+2 = 2 \)
\(|\vec b|= \sqrt{1+4+1 }=\sqrt 6 \)
\( |\vec n|=\sqrt{ 4+1+4}= \sqrt 9 = 3 \)
\( sin\ \theta = \frac{|2|}{\sqrt 6 \times 3} = \frac {2}{3\sqrt 6} \)
\(\theta = sin ^{-1}(\frac{2}{3\sqrt 6})\)
which is the point of intersection of the plane and the line

Given equation of planes are x + 2y + 3z = 2 and x-y+z+11 = 3
The Cartesian equation of a plane which passes through the line of intersection of the planes is
\(\left( { a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }z-{ d }_{ 1 } \right) +\lambda \left( { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }z-{ d }_{ 2 } \right) =0\)
\(\therefore\) The required equation of the plane is
\(\left( x+2y+3z-2 \right) +\lambda \left( x+y+z+8 \right) =0\)
\(x\left( \lambda +1 \right) +y\left( 2+\lambda \right) +z\left( 3+\lambda \right) -2+8\lambda =0\)
The distance from (3, 1, -1) to this plane is \(\frac { 2 }{ \sqrt { 3 } } \)
\(\therefore \frac { 3\left( \lambda +1 \right) +1\left( 2+\lambda \right) -1\left( 3+\lambda \right) -2+8\lambda }{ \sqrt { \left( \lambda +1 \right) ^{ 2 }+\left( 2+\lambda \right) ^{ 2 }+\left( 3+\lambda \right) ^{ 2 } } } =\frac { 2 }{ \sqrt { 3 } } \)

\(\Rightarrow \frac { 12\lambda }{ \sqrt { { 3\lambda }^{ 2 }+12\lambda +14 } } =\frac { 2 }{ \sqrt { 3 } } \)
Squaring on both sides
\( \frac{\lambda^{2}}{3 \lambda^{2}+4 \lambda+14}=\frac{1}{3} \)
\(3 \lambda^{2}=3 \lambda^{2}+4 \lambda+14 \)
\(4 \lambda=-14 \)
\(\lambda=\frac{-7}{2}\)
Putting
\(\lambda=\frac{-7}{2}\) in (1)
The required equation
\( (x+2 y+3 z-2)-\frac{7}{2}(x-y+z-3)=0 \)
\(2 x+4 y+6 z-4-7 x+7 y-7 z+21=0 \)
\(-5 x+11 y-z+17=0 \)
\(5 x-11 y+z-17=0\)

Comparing the two given lines with
\(\vec { r } =\vec { a } +t\vec { b } ,\vec { r } =\vec { c } +s\vec { d } \)
we have, \(\vec { a } =-\hat { i } -3\hat { j } -5\hat { k } ,\vec { b } =3\hat { i } +5\hat { j } +7\hat { k } ,\vec { c } =2\hat { i } +4\hat { j } +6\hat { k } \)  and \(\vec { d } =\hat { i } +4\hat { j } +7\hat { k } \)
We know that the two given lines are coplar, if \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )\) = 0
Here, \(\vec { b } \times \vec { d } \left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{matrix} \right| =7\hat { i } -14\hat { j } +7\hat { k } \)  and  \(\vec { c } -\vec { a } =3\hat { i } +7\hat { j } +11\hat { k } \)
Then, \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )=(3\hat { i } +7\hat { j } +11\hat { k } )(7\hat { i } -14\hat { j } +7\hat { k } )=0\)
Therefore the two given lines are coplanar. Then we find the non parametric form of vector equation of the plane containing the two given coplanar lines. We know that the plane containing the two given coplanar lines is
\((\vec { r } -\vec { a } ).(\vec { b } \times \vec { d } )\)= 0
which implies that \((\vec { r } -(-\hat { i } -3\hat { j } -5\hat { k } )).(7\hat { i } -14\hat { j } +7\hat { k } )\) = 0. 
Thus, the required non-parametric vector equation of the plane containing the two given coplanar lines is
\(\vec { r } .(\hat { i } -2\hat { j } +\hat { k } )\) = 0.

\(\frac { x-1 }{ 2 } =\frac { y+1 }{ \lambda } =\frac { z }{ 2 } \) and \(\frac { x-1 }{ 2 } =\frac { y+1 }{ \lambda } =\frac { z }{ \lambda } \)
\(\therefore \vec { a } =\hat { i } -\hat { j } ,\vec { b } =2\hat { i } +\lambda \hat { j } +2\hat { k } \)
\(\vec { c } =-\hat { i } -\hat { j } ,\vec { d } =5\hat { i } +2\hat { j } +\lambda \vec { k } \)
\(\left( \vec { c } -\vec { a } \right) =-2\hat { i } ,\)
and \(\left( \vec { b } \times \vec { d } \right) =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & \lambda & 2 \\ 5 & 2 & \lambda \end{matrix} \right| \)
= \(\hat { i } \left( { \lambda }^{ 2 }-4 \right) -\hat { j } \left( 2\lambda -10 \right) +\hat { k } \left( 4-5\lambda \right) \)
Since the given lines are co-planar,
\(\left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =0\)
\(\Rightarrow \left( -2\hat { i } \right) .\left[ \left( { \lambda }^{ 2 }-4 \right) \hat { i } -\hat { j } \left( 2\lambda -10 \right) +\hat { k } \left( 4-5\lambda \right) \right] =0\)
\(\Rightarrow -2\left( { \lambda }^{ 2 }-4 \right) =0\)
\(\Rightarrow { \lambda }^{ 2 }=4\)  \(\left[ \because -2\neq 0 \right] \)
\(\Rightarrow \lambda =\pm \sqrt { 4 } =\pm 2\)
The Cartesian equation of the plane containing the given lines is
\(\left| \begin{matrix} x-{ x }_{ 2 } & y-{ y }_{ 2 } & z-{ z }_{ 2 } \\ { b }_{ 1 } & { b }_{ 2 } & b_{ 3 } \\ { d }_{ 1 } & { d }_{ 2 } & { d }_{ 3 } \end{matrix} \right| \)
\(\Rightarrow \left| \begin{matrix} x+1 & y+1 & z \\ 2 & 2 & 2 \\ 5 & 2 & 2 \end{matrix} \right| =0\left[ \because \lambda =2 \right] \)
\(\Rightarrow \left( x+1 \right) \left( 4-4 \right) -\left( y+1 \right) \left( 4-10 \right) +z\left( 4-10 \right) =0\)
\(\Rightarrow \left( x+1 \right) \left( 0 \right) -\left( y+1 \right) \left( -6 \right) +z\left( -6 \right) =0\)
\(\Rightarrow 6\left( y+1 \right) -6z=0\)
\(\Rightarrow y+1-z=0\)
\(\Rightarrow y+z+1=0\) which is the required equation of the plane containing the given lines

\(\frac { x-1 }{ 1 } =\frac { y-2 }{ 1 } =\frac { z-3 }{ { m }^{ 2 } } \) and \(\frac { x-3 }{ 1 } =\frac { y-2 }{ { m }^{ 2 } } =\frac { z-1 }{ 2 } \) 
\(\therefore \vec { a } =\hat { i } +2\hat { j } +3\hat { k } ,\vec { b } =\hat { i } +\hat { j } +{ m }^{ 2 }\hat { k } \)
\(\vec { c } =\hat { i } +2\hat { j } +5\hat { k } ,\vec { d } =-3\hat { i } +{ m }^{ 2 }\hat { j } +2\hat { k } \)
\(\vec { c } -\vec { a } =-2\vec { k } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & { m }^{ 2 } \\ 1 & { m }^{ 2 } & 2 \end{matrix} \right| =\sqrt { 2 } \)
= \(\hat { i } \left( 4-{ m }^{ 4 } \right) -\hat { j } (2-{ m }^{ 2 })+\hat { k } \left( { m }^{ 2 }+3 \right) \)
Since the given lines are co-planar,
\(\left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =0 \)
\(\Rightarrow \left( -2\hat { k } \right) .\left[ \left( 4-m^{ 4 } \right) \hat { i } -\hat { j } \left( 2-{ m }^{ 2 } \right) +\hat { k } \left( m^{ 2 }-2 \right) \right] =0\)
\(\Rightarrow 2\left( { m }^{ 2 }-2 \right) =0\)
\(\Rightarrow { m }^{ 2 }-2=0\)
\(\Rightarrow { m }^{ 2 }=2\)
\(\Rightarrow m=\pm \sqrt { 2 } \)

Gives
\(\frac { x-2 }{ 1 } =\frac { y-3 }{ 1 } =\frac { z-4 }{ 3 } \) and \(\frac { x-1 }{ -3 } =\frac { y-4 }{ 2 } =\frac { z-5 }{ 1 } \)
\(\therefore \vec { a } =-2\hat { i } -3\hat { j } -4\hat { k } ,\vec { b } =\hat { i } +\hat { j } +3\hat { k } \)
\(\vec { c } =-\hat { i } -4\hat { j } -5\hat { k } ,\vec { d } =-3\hat { i } +2\hat { j } +\hat { k } \)
The two given lines are co-planar
\(y\left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) \)
\(\left( \vec { c } -\vec { a } \right) =-\hat { i } +\hat { j } +\hat { k } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 1 & 3 \\ -3 & 2 & 1 \end{matrix} \right| =\sqrt { 2 } \)
= \(\hat { i } \left( 1-6 \right) -\hat { j } (1+9)+\hat { k } \left( 2+3 \right) \)
= \(-5\hat { i } -10\hat { j } +5\hat { k } \)
\(\therefore \left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =\left( -\hat { i } +\hat { j } +\hat { k } \right) .\left( -5\hat { i } -10\hat { j } +5\hat { k } \right) \)
= 5-10 + 5= 10-10 = 0
Hence, the given lines are co-planar. Its Cartesian equation is
\(\left| \begin{matrix} x-{ x }_{ 2 } & y-{ y }_{ 2 } & z-{ z }_{ 2 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { d }_{ 1 } & { d }_{ 2 } & { d }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 & y-4 & z-5 \\ 1 & 1 & 3 \\ -3 & 2 & 1 \end{matrix} \right| =0\)
\(\Rightarrow \left( x-1 \right) \left( 1-6 \right) -\left( y-6 \right) \left( 1+9 \right) +\left( z-5 \right) \left( 2+3 \right) =0\)
\(\Rightarrow \left( x-1 \right) \left( -5 \right) -\left( y-5 \right) \left( 10 \right) +\left( z-5 \right) \left( 5 \right) =0\)
\(\Rightarrow -5x+5-10y+40+5z-25=0\)
\(\Rightarrow -5x-10y+5z+20=0\)
\(\div\) -5, we get
x + 2y - z - 4 = 0 which is the equation of the plane containing the given lines

\(\vec { r } =(5\hat { i } +7\hat { j } -3\hat { k } )+s(-4\hat { i } +4\hat { j } -5\hat { k } )\) and \(\vec { r } =(8\hat { i } +4\hat { j } +5\hat { k } )+t(7\hat { i } +\hat { j } +3\hat { k } )\)
Let \(\vec { a } =5\hat { i } +7\hat { j } -3\hat { k } ,\vec { b } =4\hat { i } +4\hat { j } -5\hat { k } \)
\(\vec { c } =8\hat { i } +4\hat { j } +5\hat { k } \ and\ \vec { d } =7\hat { i } +\hat { j } +3\hat { k } \)
We know that the two given lines are co-planar
if \(\left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =0\)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 4 & 4 & -5 \\ 7 & 1 & 3 \end{matrix} \right| \)
= \(\hat { i } \left( 12+5 \right) -\hat { j } \left( 12+35 \right) +\hat { k } \left( 4-28 \right) \)
= \(17\hat { i } -47\hat { j } -24\hat { k } \)
\(\left( \vec { c } -\vec { a } \right) =\left( 8-5 \right) \hat { i } +\left( 4-7 \right) \hat { j } +\left( 5+3 \right) \hat { k } \)
= \(3\hat { i } -3\hat { j } +8\hat { k } \)
Now,\(\left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =\left( 3\hat { i } -3\hat { j } +8\hat { k } \right) \)
\(\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) \)
= \(51+141-192=192-192=0\)
\(\therefore\) The two given lines are co-planar.
The plane containing the two given co-planar lines is
\(\left( \vec { r } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =0\)
\(\Rightarrow \left( \vec { r } -5\hat { i } +7\hat { j } -3\hat { k } \right) \times \left( 17\hat { i } -47\hat { j } -24\hat { k } \right) =0\)
\(\Rightarrow \vec { r } .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) -\left[ \left( 5\hat { i } +7\hat { j } -3\hat { k } \right) .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) \right] =0\)
\(\Rightarrow \vec { r } .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) =\left[ 85-329+72 \right] \)
\(\Rightarrow \vec { r } .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) =-172 \)
which is the required vector equation of the plane.

Equation of the plane is
\(\vec { r } =\left( 6\hat { i } -\hat { j } +\hat { k } \right) +s\left( -\hat { i } +2\hat { j } +\hat { k } \right) +t\left( -5\hat { i } -4\hat { j } -5\hat { k } \right) \)
This is the equation of the plane passing through one point \(\vec { a } =6\hat { i } -\hat { j } +\hat { k } \) and parallel to two
vectors \(\vec { b } =-\hat { i } +2\hat { j } +\hat { k } \)  and \(\vec { c } =-5\hat { i } -4\hat { j } -5\hat { k } .\)
\(\therefore\) Non-parametric form of vector equation of the plane is
\(\left( \vec { r } -\vec { a } \right) .\left( \vec { b } \times \vec { c } \right) \)
\(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ -1 & 2 & 1 \\ -5 & -4 & -5 \end{matrix} \right| \)
= \(\hat { i } \left( -10+4 \right) -\hat { j } \left( 5+5 \right) +\hat { k } \left( 4+10 \right) \)
= \(\hat { i } \left( -6 \right) -\hat { j } \left( 10 \right) +\hat { k } \left( 14 \right) \)
\(\therefore \left[ \vec { r } -\left( -6\hat { i } -\hat { j } +k \right) \right] \left[ -6\hat { i } -10\hat { j } +14\hat { k } \right] =0\)
\(\Rightarrow \vec { r } .\left( -6\hat { i } -10\hat { j } +14\hat { k } \right) .\left( -6\hat { i } -10\hat { j } +14\hat { k } \right) \)
\(\Rightarrow \vec { r } .\left( -6\hat { i } -10\hat { j } +14\hat { k } \right) -\left[ -36+10+14 \right] =0\)
\(\Rightarrow \vec { r } .\left( -6\hat { i } -10\hat { j } +14\hat { k } \right) \)
\(\div -2,\vec { r } .\left( 3\hat { i } +5\hat { j } -7\hat { k } \right) -6=0\)
Let \(\vec { r } =x\hat { i } +y\hat { i } +2\hat { k } \)
\(\Rightarrow \left( x\hat { i } +y\hat { j } +z\hat { k } \right) .\left( 3\hat { i } +5\hat { j } -7\hat { k } \right) = 6\) 
\(\Rightarrow 3x+5y-7z-6=0\) which is required Cartesian equation. 

The plane passing through three points namely
\(\vec { a } =3\hat { i } +6\hat { k } -2\hat { k } \),
\(\vec { b } =-\hat { i } -2\hat { j } +6\hat { k } \) and
\(\vec { c } =6\hat { i } +4\hat { j } -2\hat { k } \)
Now, \(\vec { b } -\vec { a } =(-1-3)\hat { i } +(-2-6)\hat { j } +(6+2)\hat { k } \)
\(=-4\hat { i } -8\hat { j } +8\hat { k } \)
\(\vec { c } -\vec { a } =(6-3)\hat { i } +(-4-6)\hat { j } +(-2+2)\hat { k } \)
\(=3\hat { i } -10\hat { j } \)
The parametric form of vector equation of the plane passing through three points is
\(\vec { r } =\vec { a } +s(\vec { b } -\vec { a } )+t(\vec { c } -\vec { a } ),s,t\in R\)
\(\Rightarrow \vec { r } =(3i+6i-2k)+s(4\hat { i } -8\hat { j } +8\hat { k } )+(3\hat { i } -10\hat { j } )s,t\in R\)
The parametric form of vector equation of the plane passing through three points is
\([\vec { r } -\vec { a } ,\vec { b } -\vec { a } ,\vec { c } -\vec { a } ]\) = 0
\(\Rightarrow \vec { r } -(3i+6i-2k).[(4\hat { i } -8\hat { j } 8\hat { k } )+(3\hat { i } -10\hat { j } )]=0\)
Cartesian equation is 
\( \Rightarrow \vec{r}(2 \hat{i}+3 \hat{j}+4 \hat{k})=16 \)
\( \Rightarrow 2 x+3 y+4 z-16=0 \)

The plane containing the line
\(\vec { r } =(\hat { i } -\hat { j } +3\hat { k } )+t(2\hat { i } -\hat { j } +4\hat { k } )\)
∴ The required plane is passing through the point \(\vec { a } =\hat { i } -\hat { j } +3\hat { k } \) and parallel to a vector \(\vec { b } =2\hat { i } -\hat { j } +4\hat { k } \) Also, the plane is perpendicular to the plane 
\(\vec { c } =\hat { i } +2\hat { j } +\hat { k } \) 
∴ The parametric form of vector equation of the plane passing through one point (\(\vec { a } \)) and parallel to two vectors \(\vec { b } \) and \(\vec { c } \) 
\(\vec { r } =\vec { a } +s\vec { b } +t\vec { c } \) where s, t ∈ R
⇒ \(\vec { r } .(\hat { i } -\hat { j } +3\hat { k } )+s(2\hat { i } -\hat { j } +4\hat { k } )+t(\hat { i } +2\hat { j } +\hat { k } )\) s, t  ∈ R 
Cartesian equation is
\(\left| \begin{matrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 & y+1 & z-3 \\ 2 & -1 & 4 \\ 1 & 2 & 1 \end{matrix} \right| =0\)
⇒ (x - 1)(-1 - 8) - (y + 1)(2 - 4) + (z - 3)(4 + 1) = 0
⇒ (x - 1)(-9) - (y + 1)(-2) + (z - 3)5 = 0
⇒ -9x + 9 +2y + 2 + 5z - 15 = 0
⇒ -9x + 2y + 5z - 4 = 0
⇒ 9x - 2y - 5z + 4 = 0

Equation of the plane passing through the point
\(=\vec { a } =\hat { i } -2\hat { j } +4 \) .......(1)
Equation of the given plane is x + 2y - 3z = 11
\(\Rightarrow \vec { r } .(\hat { i } +2\hat { j } -3\hat { k } )=11\)
The given plane is perpendicular to the vector \(\hat { i } +2\hat { j } -3\hat { k } \)
∴ The required plane is parallel to the vector 
\(\vec { b } =\hat { i } +2\hat { j } -3\hat { k } \).......(2)
The given plane is parallel to the line 
\(\frac { x+7 }{ 3 } =\frac { y+3 }{ -1 } =\frac { z }{ 1 } \) Whose direction ratios are  3,-1,1
The required plane is parallel to the vector
 \(\vec { c } =3\hat { i } -\hat { j } +\hat { k } \)     (3)
∴  The non-parametric vector equation of the plane passing through a point (\(\vec { a } \)) and parallel to two vectors \(\vec { b } \) and \(\vec { c } \) is
\((\vec { r } -\vec { a } ).(\vec { b } \times \vec { c } )=0\)
\(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & -3 \\ 3 & -1 & 1 \end{matrix} \right| \)
\(=\hat { i } (2-3)-\hat { j } (1+9)+\hat { k } (-1-6)\)
\(=-\hat { i } -10\hat { j } -7\hat { k } \)
\(\therefore (\vec { r } -(\hat { i } -2\hat { j } +4\hat { k } )).(-\hat { i } -10\hat { j } -7\hat { k } )=0\)
\([\vec { r } -(-\hat { i } -10\hat { j } -7\hat { k } )]-[(\hat { i } -2\hat { j } +4\hat { k } ).(-\hat { i } -10\hat { j } -7\hat { k } )]=0\)
\(\Rightarrow \vec { r } .(-\hat { i } -10\hat { j } -7\hat { k } )-[-1+20-28]=0\)
\(\Rightarrow \vec { r } .(-\hat { i } -10\hat { j } -7\hat { k } )\times 9=0\)
\(\Rightarrow \vec { r } .(\hat { i } +10\hat { j } +7\hat { k } )-9=0\)
\(\Rightarrow \vec { r } .(\hat { i } +10\hat { j } +7\hat { k } )=9\)
Let \(\Rightarrow \vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\therefore (x\hat { i } +y\hat { j } +z\hat { k } ).(\hat { i } +10\hat { j } +7\hat { k } )=9\)
⇒ x+10y+7z = 9 which is the required Cartesian equation of the plane.

The plane passes through two points
\(\vec { a } =2\hat { i } +2\hat { j } +\hat { k }\ and\ \vec { b } =\hat { i } -2\hat { j } +3\hat { k } \)
The straight line passing through the points
(2, 1, -3) and (-1, 5, -8) is
\(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } .\)
\(\Rightarrow \frac { x-2 }{ -1-2 } =\frac { y-1 }{ 5-1 } =\frac { z+3 }{ -8+3 } \)
\(\Rightarrow \frac { x-2 }{ -3 } =\frac { y-1 }{ 4 } =\frac { z+3 }{ -5 } \)
Hence the required plane is parallel to the vector
\(\vec { c } =-3\hat { i } +4\hat { j } -5\hat { k } \)
The parametric form of vector equation of the plane passing through two points \(\vec { a } ,\vec { b } \) parallel to a vector\(\vec { c }\ is\ \vec { r } =\vec { a } +s(\vec { b } -\vec { a } )+t\vec { c } ,s,t\in R,\)
\(\vec { r } .2\hat { i } +2\hat { j } +\hat { k } +s(-\hat { i } -4\hat { j } +2\hat { k } )+t(3\hat { i } -4\hat { j } +5\hat { k } )\) s, t ∈ R 
Cartesian form of the plane passing through two points and parallel to a vector is
\(\left| \begin{matrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { x }_{ 2 }-{ x }_{ 1 } & { y }_{ 2 }-{ y }_{ 1 } & { z }_{ 2 }-{ z }_{ 1 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| =0\)
[∵(x₁,y₁, z₁) is (2, 2, 1), (x₂, y₂, z₂) is (-1, -2, 3) & (c₁, c₂, c₃) is (-3, 4 -5)
\(\Rightarrow \left| \begin{matrix} x-2 & y-2 & z-1 \\ -1 & -4 & 2 \\ -3 & 4 & -5 \end{matrix} \right| =0\)
⇒ (x - 2)(20 - 8) - (y- 2)(5+ 6) + (z-1)(-4 -12) = 0
⇒ (x - 2)(12) - (y - 2)(11) + (z - 1)(-16) = 0
⇒ 12x- 24 -11y + 22 -16z + 16 = 0
⇒ 12x-11y-16z+14 = 0

Given plane is passing through the points
\(\vec { a } =2\hat { i } +2\hat { j } +2\hat { k }, \vec { b } =9\hat { i } +3\hat { j } +6\hat { k } \)
Equation of the given plane is 2x + 6y + 6z = 9. It can be written as \(\vec { r } .(2\hat { i } +6\hat { j } +6\hat { k } )=9\)
Since the given plane is perpendicular to \(2\hat { i } +6\hat { j } +6\hat { k } \), the required plane is parallel to \(\vec { c } =2\hat { i } +6\hat { j } +6\hat { k } \). Hence, parametric form of vector equation of plane passing through two points and parallel to a vector is
\(\vec { r } =\vec { a } +s(\vec { b } -\vec { a } )+t\vec { c } ,s,t\in R\)
\(\vec { r } =2\hat { i } +2\hat { j } +\hat { k } +s(7\hat { i } +\hat { j } +5\hat { k } )+t(2\hat { i } +6\hat { j } +6\hat { k } ),s,t\in R\)
Cartesian equation of the plane is
\(\left| \begin{matrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { x }_{ 2 }-{ x }_{ 1 } & { y }_{ 2 }-{ y }_{ 1 } & { z }_{ 2 }-{ z }_{ 1 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-2 & y-2 & z-1 \\ 7 & 1 & 5 \\ 2 & 6 & 6 \end{matrix} \right| =0\)
⇒ (x-2)(6-30) - (y-2)(42-10) + (z-1)(42-2) = 0
⇒ (x - 2)(-24) - (y - 2)(32) + (z - 1)(40) = 0
⇒ 24x + 48 - 32y + 64 + 40z - 40 = 0
⇒ -24x - 32y + 40z + 72 = 0
\(\div\) - 8 we get
3x+ 4y - 5z - 9 = 0 is the Cartesian form.
∴ The parametric form of vector equation is
\(\vec { r } =\vec { r } (3\vec { i } +4\vec { j } -5\vec { k } )=9\)

The plane passes through the point.
\(\vec { a } =2\hat { i } +3\hat { j } +6\hat { k } \) and parallel to the lines \(\frac{x-1}{2}\)
\(=\frac { y+1 }{ 3 } =\frac { z-3 }{ 1 } and\frac { x+3 }{ 2 } =\frac { y-3 }{ -5 } =\frac { z+1 }{ -3 } \)
\(\Rightarrow \vec { b } =2\hat { i } +3\hat { j } +\hat { k }\ and\ \vec { c } =2\hat { i } -5\hat { j } -3\hat { k } \)
\(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 1 \\ 2 & -5 & -3 \end{matrix} \right| \)
\(=\hat { i } (-9+5)-\hat { j } (-6-2)+\hat { k } (-10-6)\)
\(=-4\hat { i } +8\hat { j } -16\hat { k } \)
The non-parametric vector equation of the plane is
\((\vec { r } .\vec { a } ).(\vec { b } \times \vec { c } )=0,\)
\(\Rightarrow [\vec { r } (2\hat { i } +3\hat { j } +16\hat { k } ).(-4\hat { i } +8\hat { j } -16\hat { k } )]=0\)
\(\Rightarrow [\vec { r } .(-4\hat { i } +8\hat { j } -16\hat { k } )]-(-8+24-96)=0\)
\(\Rightarrow \vec { r } .(-4\hat { i } +8\hat { j } -16\hat { k } )=-80\)
\(\div -4,\) We get
\(\vec { r } .(\hat { i } -2\hat { j } +4\hat { k } )=20\)
\(Let\quad \vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\Rightarrow (x\hat { i } +y\hat { j } +z\hat { k } ).(\hat { i } -2\hat { j } +4\hat { k } )=20\)
\(\Rightarrow x=2y+4z=20\)
\(\Rightarrow x-2y+4z-20=0\)

The required plane is parallel to the given line and so it is parallel to the vector \(\vec { c } =\hat { i } +\hat { j } -\hat { k } \) and the plane passes through the points \(\vec { a } =-\hat { i } +2\hat { j } ,\vec { b } =2\hat { i } +2\hat { j } -\hat { k } \)
(i) vector equation of the plane in parametric form is \(\vec { r } =\vec { a } +s(\vec { b } -\vec { a } )+t\vec { c } \), where s, t ∈ R
which implies that \(\vec { r } =(-\hat { i } +2\hat { j } )+s(3\hat { i } -\hat { k } )+t(\hat { i } +\hat { j } -\hat { k } )\), where s, t ∈ R
(ii) vector equation of the plane in non-parametric form is \((\vec { r } -\vec { a } ).(\vec { b } -\vec { a } )\times \vec { c } )\) = 0
Now, \((\vec { b } -\vec { a } )\times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 0 & -1 \\ 1 & -1 & -1 \end{matrix} \right| =\hat { i } +2\hat { j } +3\hat { k } \)
we have \((\vec { r } -(-\hat { i } +2\hat { j } ).(\hat { i } +2\hat { j } +3\hat { k } )\) = 0 ⇒ \(\vec { r } .(\hat { i } +2\hat { j } +3\hat { k } )\) = 3
If  \(\vec { r } .(\hat { i } +2\hat { j } +3\hat { k } )\) is the position vector of an arbitrary point on the plane, then from the above equation, we get the Cartesian equation of the plane as x + 2y + 3z = 3

Given equation of line is

\(\frac { x+1 }{ 2 } =\frac { y-3 }{ 3 } =\frac { z-1 }{ -1 } \) \(\Rightarrow\) A is (-1, 3, 1) and 
\(\vec { b } =2\hat { i } +3\hat { j } -\hat { k } \)
Let  \(\frac { x+1 }{ 2 } =\frac { y-3 }{ 3 } =\frac { z-1 }{ -1 } =t\) 
If F is the foot of the ⊥ from D to the line, then it is of the form
(2t-1,3t+3,-t+1) .............(1)
\(\Rightarrow \vec { OF } =(2t-1)\hat { i } +(3t+3)\hat { j } +(-t+1)\hat { k } \)
Given point is D(5, 4, 2)
\(\Rightarrow \vec { OD } =5\hat { i } +4\hat { j } +2\hat { k } \)
\(\therefore \vec { DF } =\vec { OF } -\vec { OD } \)
\(=(2t-1-5)\hat { i } (3t+3-4)\hat { j } +(-t+1-2)\hat { k } \) .......(2)
since \(\\ \vec { b } \bot \vec { DF } ,\)we have
\(\vec { b } .\vec { DF } =0\)
⇒2(2t-6)+3(3t-1)-1(-t-1) = 0
⇒ 4t-12+9t-3+t+1 = 0
⇒ 14t -14 = 0
⇒ 14t = 14
⇒ t = 1
From(1), F is (2(1_-1,3(1)+3,-1+1) = (1,6,0)
∴ The foot of the perpendicular is (1, 6, 0)
∴ Equation of the perpendicular DF is the equation of the line passing through two points (5, 4, 2) and (1, 6, 0). Its Cartesian equation is
\(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
\(\Rightarrow \frac { x-5 }{ 1-5 } =\frac { y-4 }{ 6-4 } =\frac { z-2 }{ 0-2 } \)
\([\because ({ x }_{ 1 },y_{ 1 },{ z }_{ 1 })is(5,4,2){ (x }_{ 2 },{ y }_{ 2 },{ z }_{ 2 })is(1,6,0)]\)
\(\Rightarrow \frac { x-5 }{ -4 } =\frac { y-4 }{ 6-4 } =\frac { z-2 }{ 0-2 } \) is the equation of required perpendicular.

Given point is \(\vec { a } =-\hat { i } +2\hat { j } +\hat { k } \)
and parallel vector is \(\vec { b } =\hat { i } -2\hat { j } +\hat { k } \)
\(\therefore\) The parametric form of vector equation of a line passing through \(\vec { a} \) and parallel to \(\vec { b } \) is 
\(\vec { r } =\vec { a } +\vec { b } \)
\(\vec { r } =(-\hat { i } +2\hat { j } +\hat { k } )+t(\hat { i } -2\hat { j } +\hat { k } ),k\in R\)  (1)
Given line is 
\(\vec { r } =(2\hat { i } +3\hat { j } -\hat { k } )+t(\hat { i } -2\hat { j } +\hat { k } ),k\in R\quad (2)\)
\(\therefore \vec { c } =2\hat { i } +3\hat { j } -\hat { k } \)
\(\therefore \vec { c } -\vec { a } =(2\hat { i } +3\hat { j } -\hat { k } )-(\hat { i } -2\hat { j } +\hat { k } )\)
\(=(3\hat { i } +\hat { j } -2\hat { k } )\)
\((\vec { c } -\vec { a } )\times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & \frac { 1 }{ 2 } & -\frac { 1 }{ 12 } \\ 1 & 1 & \frac { 1 }{ 6 } \end{matrix} \right| \)
\(=\hat { i } (1-4)-\hat { j } (3+2)+\hat { k } (-6-1)\)
\(=-3\hat { i } -5\hat { j } -7\hat { k } \)
\(|(\vec { c } -\vec { a } )\times \vec { b } |=\sqrt { { (-3) }^{ 2 }+{ (-5) }^{ 2 }+{ (-7) }^{ 2 } } \)
\(=\sqrt { 9+25+49 } =\sqrt { 83 } \)
\(|\vec { b } |=\sqrt { { 1 }^{ 2 }+{ (-2) }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 6 } \)
\(\therefore\) Distance between the parallel lines
\(d=\frac { |(\vec { c } -\vec { a } )\times \vec { b } | }{ |\vec { b } | } =\frac { \sqrt { 83 } }{ \sqrt { 6 } } \) units

Given lines are x+1 = 2y = -12z
\(\Rightarrow \frac { x+1 }{ 1 } =\frac { y-0 }{ 1 } =\frac { z-0 }{ \frac { -1 }{ 12 } } \)
and x = y + 2 = 6z - 6
\(\Rightarrow \frac { x-0 }{ 1 } =\frac { y+2 }{ 1 } =\frac { z-1 }{ \frac { 1 }{ 6 } } \)
\(\therefore \vec { a } =-\hat { i } ,\vec { b } =\hat { i } +\frac { 1 }{ 2 } \vec { j } -\frac { 1 }{ 12 } \hat { k } \)
\(\vec { c } =-2\hat { j } +\hat { k } \ and\ \vec { d } =\hat { i } +\hat { j } +\frac { 1 }{ 6 } \hat { k } \)
\(\vec { c } -\vec { a } =2\hat { j } +\hat { k } -(-\hat { i } )=\hat { i } -2\hat { j } +\hat { k } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & \frac { 1 }{ 2 } & -\frac { 1 }{ 12 } \\ 1 & 1 & \frac { 1 }{ 6 } \end{matrix} \right| \)
\(=\hat { i } \left( \frac { 1 }{ 12 } +\frac { 1 }{ 12 } \right) -\hat { j } \left( \frac { 1 }{ 6 } +\frac { 1 }{ 12 } \right) +\hat { k } \left( 1-\frac { 1 }{ 12 } \right) \)
\(=\frac { 1 }{ 6 } \hat { i } -\frac { 1 }{ 4 } \hat { j } +\frac { 1 }{ 2 } \hat { k } \)
Now \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )\)
\(=\left( \hat { i } -2\hat { j } +\hat { k } \right) .\left( \frac { 1 }{ 6 } \hat { i } -\frac { 1 }{ 4 } \hat { j } +\frac { 1 }{ 2 } \hat { k } \right) \)
\(=\frac { 1 }{ 6 } +\frac { 2 }{ 4 } +\frac { 1 }{ 2 } =\frac { 1 }{ 6 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } \)
\(=\frac { 1 }{ 6 } +1=\frac { 7 }{ 6 } \)
Since \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )\)0, the given lines are skew lines. 
\(\therefore |\vec { b } \times \vec { d } |=\sqrt { \frac { 1 }{ 36 } +\frac { 1 }{ 16 } +\frac { 1 }{ 4 } } \)
\(\sqrt { \frac { 4+9+36 }{ 144 } } =\sqrt { \frac { 49 }{ 144 } } =\frac { 7 }{ 12 } \)
Shortest distances between the skew lines

Given lines are \(\frac { x-3 }{ 3 } =\frac { y-3 }{ -1 } \)....(1)
and z-1 = 0 
\(\Rightarrow\) z = 1
and \(\frac { x-6 }{ 2 } =\frac { z-1 }{ 3 } \) ...(2)  
and y-2 = 0\(\Rightarrow\) y = 2
Substituting y = 2 and z = 1 in (1) we get
\(\frac { x-3 }{ 3 } =\frac { 2-3 }{ -1 } =\frac { -1 }{ -1 } =1\Rightarrow x-3=3\Rightarrow x=6\)
The point of intersection is (6, 2, 1)
Let us check whether (6, 2, 1) satisfies (1) and (2)
\((1)\rightarrow \frac { 6-6 }{ 2 } =\frac { 1-1 }{ 3 } \Rightarrow 0=0\)
\((2)\rightarrow \frac { 6-3 }{ 3 } =\frac { 1-3 }{ -1 } \Rightarrow -1=-1\)
Hence, the given two lines intersect and the point of intersection is (6, 2, 1).

Given lines are
\(\vec { r } =(6\hat { i } +\hat { j } +2\hat { k } )+s(\hat { i } +2\hat { j } -3\hat { k } )\)
\(\vec { a } =6\hat { i } +\hat { j } +2\hat { k } ,\vec { b } =\hat { i } +2\hat { j } -3\hat { k } \)
and \(\vec { r } =(3\hat { i } +2\hat { j } -2\hat { k } )+t(2\hat { i } +4\hat { j } -5\hat { k } )\)
\(\vec { c } =3\hat { i } +2\hat { j } -2\hat { k } \quad and\quad \vec d = 2\hat { i } +4\hat { j } -5\hat { k } \)
Since \(\vec { b } \neq \vec { d } \), they are not parallel and they do not intersect.
Hence the given lines are skew lines.
Shortest distance between the two skew lines
\(\delta =\frac { |(\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )| }{ |(\vec { b } \times \vec { d } )| } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{matrix} \right| \)
\(=\hat { i } (-10+12)-\hat { j } (-5+6)+\hat { k } (4-4)\)
\(=2\hat { i } -\hat { j } \Rightarrow |\vec { b } \times \vec { d } |\quad \sqrt { { 2 }^{ 2 }++(-1)^{ 2 } } =\sqrt { 5 } \)
\(\vec { c } =\vec { a } =(3\hat { i } +2\hat { j } -2\hat { k } )-(6\hat { i } +\hat { j } +2\hat { k } )\)
\(=-3\hat { i } +\hat { j } -4\hat { k } \)
\(\therefore (\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )=(-3\hat { i } +\hat { j } -4\hat { k } ).(2\hat { i } -\hat { j } )\)
= -6-1 = -7
\(\therefore \delta =\frac { |-7| }{ \sqrt { 5 } } =\frac { 7 }{ \sqrt { 5 } } units\)

Comparing the given equation \(\vec { r } =(\hat { i } -4\hat { j } +3\hat { k } )+t(2\hat { i } +3\hat { j } +\hat { k } )\) with \(\vec { r } =\hat { a } +t\hat { b } \),
we get \(\vec { a } =\hat { i } -4\hat { j } +3\hat { k } \) and \(\vec { b } =2\hat { i } +3\hat { j } +\hat { k } \). We denote the given point (-1, 2, 3) by D and the point (1, -4, 3) on the straight line by F.
If F is the foot of the perpendicular from to the straight line, then F is of the form (2t + 1, 3t - 4, t+3)  and \(\vec { DF } =\vec { OF } -\vec { OD } =(2t+2)\hat { i } +(3t-6)\hat { j } +t\hat { k } \)

Since \(\vec { b } \) is perpendicular to \(\vec { DF } \), we have
\(\vec { b } .\vec { DF } \) = 0 ⇒ 2(2t + 2)+ 3(3t - 6) + 1(t) ⇒ t= 1
Therefore, the coordinate of F is (3, -1, 4)
Now, the perpendicular distance from the given point to the given line is
\(DF=\left| \vec { DF } \right| =\sqrt { { 4 }^{ 2 }+(-{ 3) }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 26 } \)

The straight line passes through the points (-5, 7, -4) and (13, -5, 2), and therefore, direction ratios of the straight line joining these two points are 18, -12, 6. That is 3, -2, 1.
So, the straight line is parallel to \(3\hat { i } -2\hat { j } +\hat { k } \). Therefore,
(i) Required vector equation of the straight line in parametric form is \(\vec { r } =(-5\hat { i } +7\hat { j } -4\hat { k } )+t(3\hat { i } -2\hat { j } +\hat { k } )\)or \(\vec { r } =(13\hat { i } -5\hat { j } +2\hat { k } )+s(3\hat { i } -2\hat { j } +\hat { k } )\) where s, t ∈ R.
(ii) Required Cartesian equations of the straight line are \(\frac { x+5 }{ 3 } =\frac { y-7 }{ -2 } =\frac { z+4 }{ 1 } \) or \(\frac { x-13 }{ 3 } =\frac { y+5 }{ -2 } =\frac { z-2 }{ 1 } \)
An arbitrary point on the straight line is of the form
(3t - 5, 2t + 7, t - 4) or (3s + 13, -2s - 5, s + 2)
Since the straight line crosses the xy-plane, the z-coordinate of the point of intersection is zero.
Therefore, we have t - 4 = 0, that is, t - 4, and hence the straight line crosses the xy-plane at (7, -1, 0)

Comparing the given equation with equation of a straight line \(\vec { r } =\vec { a } +t\vec { b } \), we have \(\vec { a } =3\hat { i } -2\hat { j } +6\hat { k } \) and \(\vec { b } =3\hat { i } -2\hat { j } +6\hat { k } \). Therefore,
(i) If \(\vec { b } ={ b }_{ 1 }\hat { i } +{ b }_{ 2 }\hat { j } +{ b }_{ 3 }\hat { k } \), then direction ratios of the straight line are \({ b }_{ 1 },{ b }_{ 2 },{ b }_{ 3 }\). Therefore, direction ratios of the given straight line are proportional to 2, −1,3, and hence the direction cosines of the given straight line are \(\frac { 2 }{ \sqrt { 14 } } ,\frac { -1 }{ \sqrt { 14 } } ,\frac { 3 }{ \sqrt { 14 } } \)
(ii) vector equation of the straight line in non-parametric form is given by \((\vec { r } -\vec { a } )\times \vec { b } =\vec { 0 } \), Therefore, \((\vec{r}-(3\hat { i } -2\hat { j } +6\hat { k } )\times(2\hat { i } -\hat { j } +3\hat { k } )=\vec{0}\)
(iii) Here (x₁, y₁ ,z₁ ) = (3, −2,6) and the direction ratios are proportional to 2, −1,3. Therefore, Cartesian equations of the straight line are \(\frac { x-3 }{ 2 } =\frac { y+2 }{ -1 } =\frac { z-6 }{ 3 } \)

The required line passes through (1, 2, −3). So, the position vector of the point is \(\hat { i } +2\hat { j } -3\hat { k } \).
Let \(\vec { a } =\hat { i } +2\hat { j } -3\hat { k } \) and \(\vec { b } =4\hat { i } +5\hat { j } -7\hat { k } \). Then, we have
(i) vector equation of the required straight line in parametric form is \(\vec { r } =\vec { a } +t\vec { b } \), t ∈ R
Therefore, \(\vec { r } =(\hat { i } +2\hat { j } -3\hat { k } )+t(4\hat { i } +5\hat { j } -7\hat { k } )\), t ∈ R
(ii) vector equation of the required straight line in non-parametric form is \((\vec { r } -\vec { a } )\times \vec { b } =\vec { 0 } \)
Therefore, \((\vec { r } -(\hat { i } +2\hat { j } -3\hat { k } ))\times t(4\hat { i } +5\hat { j } -7\hat { k } )=\vec { 0 } \)
(iii) Cartesian equations of the required line are \(\frac { x{ -x }_{ 1 } }{ { b }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { b }_{ 2 } } =\frac { z{ -z }_{ 1 } }{ { b }_{ 3 } } \)
Here, (x₁, y₁, z₁ ) = (1, 2, −3) and direction ratios of the required line are proportional to 4, 5, −7. Therefore, Cartesian equations of the straight line are \(\frac { x-1 }{ 4 } =\frac { y-2 }{ 5 } =\frac { z+3 }{ -7 } \)

Given \(\vec { a } =2\hat { i } +3\hat { j } -\hat { k } ,\vec { b } =-\hat { i } +2\hat { j } -4\hat { k } ,\vec { c } =\hat { i } +\hat { j } +\hat { k } \)
\(\vec { a } \times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & -1 \\ -1 & 2 & -4 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} 3 & -1 \\ 2 & -4 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 2 & -1 \\ -1 & -4 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 3 \\ -1 & 2 \end{matrix} \right| \)
= \(\hat { i } (-12+2)-\hat { j } (-8-1)+\hat { k } (4+3)\)
= \(-10\hat { i } +9\hat { j } +7\hat { k } \)
\(\vec { a } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & -1 \\ 1 & 1 & 1 \end{matrix} \right| =\hat { i } \left| \begin{matrix} 3 & -1 \\ 1 & 1 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \)
= \(\hat { i } (3+1)-\hat { j } (2+1)+\hat { k } (2-3)\)
= \(4\hat { i } -3\hat { j } -\hat { k } \)
∴ \((\vec { a } \times \vec { b } ).(\vec { a } \times \vec { c } )=(-10\hat { i } +9\hat { j } +7\hat { k } ).(4\hat { i } -3\hat { j } -\hat { k } )\)
= -40-27-7 = -74

  Given \(\vec { a } =2\hat { i } +3\hat { j } -\hat { k } ,\vec { b } =3\hat { i } +5\hat { j } +2\hat { k } \) and \(\vec { c } =-\hat { i } -2\hat { j } +3\hat { k } \)
Consider \((\vec { a } \times \vec { b } )=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & -1 \\ 3 & 5 & 2 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} 3 & -1 \\ 5 & 2 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 3 \\ 3 & 5 \end{matrix} \right| \)
= \(\\ \hat { i } (6+5)-\hat { j } (4+3)+\hat { k } (10-9)=11\hat { i } -7\hat { j } +\hat { k } \)
∴ LHS = \((\vec { a } \times \vec { b } )\times \vec { c } \)
= \(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & -1 \\ 3 & 5 & 2 \end{matrix} \right| =\hat { i } \left| \begin{matrix} -7 & 1 \\ -2 & 3 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 11 & 1 \\ -1 & 3 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 11 & -7 \\ -1 & -2 \end{matrix} \right| \)
= \(\hat { i } (-21+2)+\hat { j } (33+1)+\hat { k } (-22-7)\)
= \(-19\hat { i } -34\hat { j } -29\hat { k } \) ..............(1)
For RHS
\(\vec { a } .\vec { i } =(2\hat { i } +3\hat { j } -\hat { k } ).(-\hat { i } -2\hat { j } +3\hat { k } )\)
= -2-6-3 = -11
\(\vec { b } .\vec { c } =(3\hat { i } +5\hat { j } +2\hat { k } ).(-\hat { i } -2\hat { j } +3\hat { k } )\)
= -3-10+6 = -7
∴ RHS = \((\vec { a } .\vec { c } )\vec { b } -(\vec { b } .\vec { c } )\vec { a } \)
= \(-11(3\hat { i } +5\hat { j } +2\hat { k } )+7(2\hat { i } +3\hat { j } -\hat { k } )\)
= \(-33\hat { i } -55\hat { j } -22\hat { k } +14\hat { i } +21\hat { j } -7\hat { k } \)
= \(-19\hat { i } -34\hat { j } -29\hat { k } \)  ............... (2)
From (1) & (2), LHS = RHS
Hence \((\vec { a } \times \vec { b } )\times \vec { c } =(\vec { a } .\vec { c } )\vec { b } -(\vec { b } .\vec { c } )\vec { a } \)
(ii) \(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 5 & 2 \\ -1 & -2 & 3 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} 5 & 2 \\ -2 & 3 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 3 & 2 \\ -1 & 3 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 3 & 5 \\ -1 & -2 \end{matrix} \right| \)
= \(\hat { i } (15+4)-\hat { j } (9+2)+\hat { k } (-6+5)\)
= \(19\hat { i } -11\hat { j } -\hat { k } \)
∴ \(\vec { a } (\vec { b } \times \vec { c } )=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & -1 \\ 19 & -11 & -1 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} 3 & -1 \\ -11 & -1 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 2 & -1 \\ 19 & -1 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 3 \\ 19 & -11 \end{matrix} \right| \)
= \(\hat { i } (-3-11)-\hat { j } (-2+19)+\hat { k } (-22-57)\)
= \(-14\hat { i } -17\hat { j } -79\hat { k } \) ............(1)
For RHS
\(\vec { a } .\vec { c } =-11\Rightarrow (\vec { a } .\vec { c } )\vec { b } =-11(3\hat { i } +5\hat { j } +2\hat { k } )\)
= -\(33\hat { i } -55\hat { j } -22\hat { k } \)
\(\vec { a } .\vec { b } =(2\hat { i } +3\hat { j } -\hat { k } ).(3\hat { i } +5\hat { j } +2\hat { k } )\)
= 6+15-2 = 19
\((\vec { a } .\vec { b } )\vec { c } =19(-\hat { i } -2\hat { j } +3\hat { k } )=-19\hat { i } -38\hat { j } +57\hat { k } \)
RHS = \((\vec { a } .\vec { c } )\vec { b } -(\vec { a } .\vec { b } )\vec { c } \)
= \(-33\hat { i } -55\hat { j } -22\hat { k } -(-19\hat { i } -38\hat { j } +57\hat { k } )\)
= \(-14\hat { i } -17\hat { j } -79\hat { k } \) ............(2)
From (1) & (2), LHS = RHS
∴ \(\vec { a } \times (\vec { b } \times \vec { c } )=(\vec { a } .\vec { c } )\vec { b } -(\vec { a } .\vec { b } )\vec { c } \)

By definition,
\(\vec { a } \times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -1 & 0 \\ 1 & -1 & -4 \end{matrix} \right| =4\hat { i } +4\hat { j } ,\vec { c } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 0 & 3 & -1 \\ 2 & 5 & 1 \end{matrix} \right| =8\hat { i } -2\hat { j } -6\hat { k } \)
\((\vec { a } \times \vec { b } )(\vec { c } \times \vec { d } )=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 4 & 4 & 0 \\ 8 & -2 & -6 \end{matrix} \right| =-24\hat { i } +24\hat { j } -40\hat { k } \) ..............(1)
On the other hand, we have
\([\vec { a }, \vec { b }, \vec { d } ]\vec { c } -[\vec { a } ,\vec { b } ,\vec { c } ]\vec { d } =28(3\vec { j } -\vec { k } )-12(2\hat { i } +5\hat { j } +\hat { k } )=-24\hat { i } +24\hat { j } -40\hat { k } \) ..............(2)
Therefore, from equations (1) and (2), identity (i) is verified.
The verification of identity (ii) is left as an exercise to the reader

Let A = (6, -7, 0), B = (16, -19, -4), C = (0, 3, -6), D = (2, -5, 10). To show that the four points A, B, C, D lie on a plane,
we have to prove that the three vectors \(\vec { AB } ,\vec { AC } ,\vec { AD } \)  are coplanar.
Now, \(\vec { AB } =\vec { OB } -\vec { OA } =(16\hat { i } -19\hat { j } -4\hat { k } )-(6\hat { i } -7\hat { j } )=10\hat { i } -12\hat { j } -4\hat { k } \)
\(\vec { AC } =\vec { OC } -\vec { OA } =-6\hat { i } +10\hat { j } -6\hat { k } \) and \(\vec { AC } =\vec { OC } -\vec { OA } =-6\hat { i } +10\hat { j } -6\hat { k } \)
We have \([\vec { AB } ,\vec { AC } ,\vec { AD } ]\) = \(\left| \begin{matrix} 10 & -12 & -4 \\ -6 & 10 & -6 \\ -4 & 2 & 10 \end{matrix} \right| \) = 0
Therefore, the three vectors \(\vec { AB } ,\vec { AC } ,\vec { AD } \) are coplanar and hence the four points A, B, C and D lie on a plane

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors and which make angles α,β respectively with positive x-axis
Draw AL and BM 丄 to x-axis
Then \(|\vec { OL } |=|\vec { OA } |cos\alpha \Rightarrow \vec { OL } =\vec { |OL| } \hat { i } =cos\alpha \hat { i } \)
\(|\vec { LA } |=|\vec { OB } |\) sin α
⇒ \(\vec { LA } =|\vec { OB } |\hat { j } =sin\alpha (-\hat { j } )=-sin\alpha \hat { j } \)
[\(\vec { LA } \) is in the opp direction of y axis]
\(\hat { a } =\vec { OA } =\vec { OL } +\vec { LA } =cos\alpha \hat { i } -sin\alpha \check { j } \) ..(1)
Similarly \(\hat { b } =\vec { OB } =\vec { OM } +\vec { MB } =cos\beta \hat { i } +sin\beta \hat { j } \) ...(2)
Now \(\hat { a } \times \hat { b } =|\hat { a } ||\hat { b } |sin(\alpha +\beta )\hat { k } =sin(\alpha +\beta )\hat { k } \) ....(3)
[\(|\hat { a } |=|\hat { b } |\) = 1]
Also \(\hat { a } \times \hat { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ cos\alpha & -sin\alpha & 0 \\ cos\beta & cos\beta & 0 \end{matrix} \right| \)
= \(\hat { i } (0)-\hat { j } (0)+\hat { k } \)(cosα sinβ + sinα cosβ)
= (sin α cos β + cos α sin β)\(\hat { k } \) .(4)
using (3) and (4), sin(α+β) = sin α cos β + cos α sin β

Let the position vector of the vertices of ΔABC be  \(\vec { a } \), \(\vec { b } \) and  \(\vec { c } \) respectively.
Since G is the centroid of ΔABC, \(\vec { OG } =\frac { \vec { a } +\vec { b } +\vec { c } }{ 3 } \)
Are of ΔGAB, = \(|\vec { AB } \times \vec { AG } |=|(\vec { OB } -\vec { OA } )\times (\vec { OG } -\vec { OA } )|\)
= \(\left| (\vec { b } -\vec { a } )\times \left( \frac { \vec { a } +\vec { b } +\vec { c } }{ 3 } -\vec { a } \right) \right| \)
= \(\left| (\vec { b } -\vec { a } )\times \left( \frac { \vec { a } +\vec { b } +\vec { c } -3\vec { a } }{ 3 } \right) \right| \)
= \(\left| (\vec { b } -\vec { a } )\times \left( \frac { \vec { b } +\vec { c } +2\vec { a } }{ 3 } \right) \right| \)
= \(\frac { 1 }{ 3 } |(\vec { b } -\vec { a } )\times (\vec { a } +\vec { 0 } -2\vec { a } )|\)
= \(\frac { 1 }{ 3 } |\vec { b } \times \vec { b } +\vec { b } \times \vec { c } -2\vec { b } \times \vec { a } -\vec { a } \times \vec { b } -\vec { a } \times \vec { c } +2\vec { a } \times \vec { a } |\)
[∵ cross product is distributive]
= \(\frac { 1 }{ 3 } |\vec { b } \times \vec { c } +2\vec { a } \times \vec { b } -\vec { a } \times \vec { b } +\vec { c } \times \vec { a } |\)
= \(\frac { 1 }{ 3 } |\vec { a } \times \vec { b } +\vec { b } \times \vec { c } +\vec { c } \times \vec { a } |\)  \( [\because \vec { b } \times \vec { b } =\vec { 0 } ,\vec { a } \times \vec { a } =\vec { 0 } ,.(1)\vec { a } \times \vec { b } =-\vec { b } \times \vec { a } ]\)
Area of ΔGAC = \(|\vec { CA } \times \vec { AG } |\)
= \(|(\vec { OA } -\vec { OC } )\times (\vec { OG } -\vec { OA } )|\)
\(\left| (\vec { a } -\vec { c } )\times \left( \frac { \vec { a } +\vec { b } +\vec { c } }{ 3 } -\vec { a } \right) \right| \)
= \(\left| (\vec { c } -\vec { a } )\times \left( \frac { \vec { a } +\vec { b } +\vec { c } -3\vec { a } }{ 3 } \right) \right| \)
= \(\frac { 1 }{ 3 } |(\vec { a } -\vec { c } )\times (\vec { b } +\vec { c } -2\vec { a } )|\)
= \(\frac { 1 }{ 3 } |\vec { a } \times \vec { b } +\vec { a } \times \vec { c } -2\vec { a } \times \vec { a } -\vec { c } \times \vec { b } -\vec { c } \times \vec { c } +2\vec { c } \times \vec { a } |\)
= \(\frac { 1 }{ 3 } |\vec { a } \times \vec { b } -\vec { c } \times \vec { a } +\vec { b } \times \vec { c } +2\vec { c } \times \vec { a } |\)
= \(\frac { 1 }{ 3 } |\vec { a } \times \vec { b } +\vec { b } \times \vec { c } +\vec { c } \times \vec { a } |\)
Also area of ΔGBC = \(|\vec { BC } \times \vec { BG } |\)
= \(|\vec { OC } -\vec { OB } )\times (\vec { OG } -\vec { OB } )|\)
= \(\left| (\vec { c } -\vec { a } )\times \left( \frac { \vec { a } +\vec { b } +\vec { c } -\vec { b } }{ 3 } \right) \right| \)
= \(\left| (\vec { c } -\vec { a } )\times \left( \frac { \vec { a } +\vec { b } -2\vec { a } }{ 3 } \right) \right| \)
\(\frac { 1 }{ 3 } |(\vec { c } -\vec { b } )\times (\vec { a } +\vec { c } -2\vec { b } )|\) 
= \(\frac { 1 }{ 3 } |\vec { c } \times \vec { a } +\vec { c } \times \vec { c } -2\vec { c } \times \vec { b } -\vec { b } \times \vec { a } -\vec { b } \times \vec { c } +2\vec { b } \times \vec { b } |\)
= \(\frac { 1 }{ 3 } |\vec { c } \times \vec { a } +2\vec { b } \times \vec { c } +\vec { a } \times \vec { b } -\vec { b } \times \vec { c } |\)
\(\frac { 1 }{ 3 } |\vec { c } \times \vec { a } +\vec { b } \times \vec { c } +\vec { a } \times \vec { b } |\)
From (1), (2) and (3),
Area of ΔGAB = Area of ΔGAC = Area of ΔGBC
= \(\frac { 1 }{ 3 } |\vec { a } \times \vec { b } +\vec { b } \times \vec { c } +\vec { c } \times \vec { a } |\)
= \(\frac { 1 }{ 3 } \) Area of ΔABC.

In triangle ABC, consider A as the origin. Then the position vectors of D, E, F are given by \(\frac { \vec { AB } +\vec { AC } }{ 2 } ,\frac { \vec { AC } }{ 2 } ,\frac { \vec { AB } }{ 2 } \) respectively.
Since \(​​\left| \vec { AB } \times \vec { AC } \right| \) is the area of the parallelogram formed by the two vectors \(\vec { AB }\), \(\vec { AC } \) as adjacent sides, the area of  ΔABC is \(\frac{1}{2}\) \(​​\left| \vec { AB } \times \vec { AC } \right| \). Similarly, considering ΔDEF, we get

the area of ΔDEF =  \(\frac{1}{2}\)  \(​​\left| \vec { DE } \times \vec { DF } \right| \)
= \(\frac{1}{2}\) \(​​\left| (\vec { AE }-\vec{AD}) \times (\vec { AF }-\vec{AD}) \right|​​\)
= \(\left| \frac { \vec { AB } }{ 2 } \times \frac { \vec { AC } }{ 2 } \right| \)
= \(\frac14\) \(\left( \frac { 1 }{ 2 } \left| \vec { AB } \times \vec { AC } \right| \right) \)
= \(\frac14\)(the area of ΔABC)

Consider a triangle ABC in which the two altitudes AD and BE intersect at O. Let CO be produced to meet AB at F. We take O as the origin and let \(\vec { OA } =\vec { a } \), \(\vec { OB } =\vec { b} \) and  \(\vec { OC } =\vec { c } \)

Since \(\vec { AD } \) is perpendicular to \(\vec { BC } \), we have \(\vec { OA } \) is perpendicular to \(\vec { BC } \), and
hence we get \(\vec { OA } \) . \(\vec { BC } \) = 0. That is, \(\vec { a } .(\vec { c } -\vec { b } )=0\), which means
\(\vec { a } .\hat{c}-\hat{a}.\hat{b}=0\)....(1)
Similarly, since \(\vec { BE } \) is perpendicular to \(\vec { CA } \), we have \(\vec { OB } \) is perpendicular to \(\vec { CA } \), and hence we get \(\vec { OB } .\vec { CA } \) = 0.
That is, \(\vec {b } .(\vec {a } -\vec { c } )=0\)
\(\vec { a } .\hat{c}-\hat{b}.\hat{c}=0\).......(2)
Adding equations (1) and (2), gives  \(\vec { a } .\hat{c}-\hat{b}.\hat{c}=0\). That is, \(\hat{c}(\hat{a}-\hat{b})=0\)
That is \(\vec { OC } \) . \(\vec { BA } \) = 0.
Therefore, \(\vec { BA } \) is perpendicular to \(\vec { OC} \). 
Which implies that \(\vec { CF} \) is perpendicular to \(\vec { AB } \). 
Hence, the perpendicular drawn from C to the side AB passes through O. Therefore, the altitudes are concurrent.

Let A be the origin, \(\vec { b } \) be the position vector of B and \(\vec {c } \) be the position vector of C .
Now D is the midpoint of BC , and so the position vector of D \(\frac{\vec{b}+\vec{c}}{2}\). There, we get

\({ \left| \vec { AD } \right| }^{ 2 }=\vec { AD } .\vec { AD } \)= \(\left( \frac { \vec { b } +\vec { c } }{ 2 } \right) .\left( \frac { \vec { b } +\vec { c } }{ 2 } \right) \)= \(\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }+2\overrightarrow { b } .\vec { c } )\)....(1)
Now, \(\vec { BD } =\vec { AD } -\vec { AB } \) = \(\frac { \vec { b } +\vec { c } }{ 2 } -\vec { b }=\frac { \vec {c } -\vec { b} }{ 2 }\)
Then, we get, =\({ \left| \vec { BD } \right| }^{ 2 }=\vec { BD } .\vec {BD } \) = \(\left( \frac { \vec { c } -\vec { b} }{ 2 } \right) .\left( \frac { \vec { c } -\vec { b} }{ 2 } \right) \) = \(\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }-2\overrightarrow { b } .\vec { c } )\)....(2)
Now, adding (1) and (2), we get
Therefore, \({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 }=\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }+2\overrightarrow { b } .\vec { c } )+\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }-2\overrightarrow { b } .\vec { c } )=\frac { 1 }{ 2 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 })\)
⇒ \({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 }=\frac { 1 }{ 2 } ({ \left| \vec { AB } \right| }^{ 2 }+{ \left| \vec { AC } \right| }^{ 2 })\)
Hence, \({ \left| \vec { AB } \right| }^{ 2 }+{ \left| \vec { AC } \right| }^{ 2 }=2({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 })\)

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors and which make angles α and β, respectively, with positive x-axis, where A and B are as in the diagram.
Draw AL and BM perpendicular to the x-axis. Then \(\left| \vec { OL } \right| =\left| \vec { OA } \right| \) cos α = cos α, \(\left| \vec { LA } \right| =\left| \vec { OA } \right| \) sin α =  sin α
So, \(\vec { OL } =\left| \vec { OL } \right| \)\(\hat { i } \) =  cos,α \(\hat { i } \), \(\overrightarrow { LA } \) = sin α (-\(\hat { j } \))
Therefore, \(\hat { a } =\overrightarrow { OA} = \overrightarrow { OL } +\overrightarrow { LA } \) = cos α \(\hat { i } \) - sin α \(\hat { j } \) ..(1)
Similarly \(\hat { b } \) = cos β \(\hat { i } \)+ sin β \(\hat { j } \) ....(2)
The angle between \(\hat { a } \) and \(\hat{b}\) is α + β and so,
\(\hat { a } .\hat { b } =\left| \hat { a } \right| \left| \hat { b } \right| \) cos (α + β) = cos (α + β) ... (3)

On the other hand, from (1) and (2)
\(\hat { a } .\hat { b } =(cos\alpha \hat { i } -sina\hat { j } )(cos\beta \hat { i } -sin\beta \hat { j } )\) = cos α cos β - sin α sin β....(4)
From (3) and (4), we get cos(α + β) = cos α cos β - sin α sin β

Equation of the given plane is 2x - 3y + 5z +7 = 0
Equation of the plane parallel to the given plane
is 2x - 3y + 5z + k = 0  ..(1)
Since this plane passes through the point (3, 4, -1). we get
2(3) - 3(4) + 5(-1) + k = 0
\(\Rightarrow\) 6 - 12 - 5 + k = 0
\(\Rightarrow\) 11 +k = 0  

k = 11
\(\therefore\) (1) becomes, 2x - 3y + 5z + 11 = 0 which is the equation of the required plane. Distance between two parallel planes.
= \(\frac { \left| { d }_{ 1 }-{ d }_{ 2 } \right| }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } \)
\(d=\frac { \left| 7-11 \right| }{ \sqrt { { 2 }^{ 2 }+\left( -3 \right) ^{ 2 }+{ 5 }^{ 2 } } } \)
= \(\cfrac { \left| -4 \right| }{ \sqrt { 4+9+25 } } \)
= \(\cfrac { 4 }{ \sqrt { 38 } } \) units.

\(\vec { r } .(2\hat { i } -7\hat { j } +4\hat { k } )=3\) and 3x - 5y + 11 = 0,
The vector equation of a plane passing through the line of intersection of the planes
\(\vec { r } .\vec { { n }_{ 1 } } ={ d }_{ 1 }\) and \(\vec { r } .\vec { n_{ 2 } } ={ d }_{ 2 }\) is given by
\(\left( \vec { r } .\vec { { n }_{ 1 } } -{ d }_{ 1 } \right) +\lambda \left( \vec { r } .{ \vec { n } }_{ 2 }-{ d }_{ 2 } \right) =0\) ...(1)
put \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\({ \vec { n } }_{ 1 }=2\hat { i } -7\hat { j } +4\hat { k } ,{ \vec { n } }_{ 2 }=3\hat { i } -5\hat { j } +4\hat { k } \)
\({ d }_{ 1 }=+3,{ d }_{ 2 }=-11\) in (1) we get
\(\left[ (x\hat { i } +y\hat { j } +z\hat { k } ).\left( 2\hat { i } -7\hat { j } +4\hat { k } -3 \right) \right] +\lambda \left[ \left( x\hat { i } +y\hat { j } +z\hat { k } \right) .\left( 3\hat { i } -5y+4z+11 \right) \right] =0\)
\(\Rightarrow \left( 2x-7y+4z-3 \right) +\lambda \left( 3x-5y+4z+11 \right) =0\)....2
Since the plane passes through the point (-2, 1, 3) we get,
\(\left[ 2(-2)-7(1)+4(3)-3 \right] +\lambda \left[ -6-5+12+11 \right] =0\)
\(\Rightarrow (-4-7+12-3)+\lambda (12)=0\)
\(\Rightarrow -2+12\lambda =0\Rightarrow 12\lambda =2\Rightarrow \lambda =\frac { 1 }{ 6 } \)
Substituting \(\lambda =\frac { 1 }{ 6 } \) in (1) we get,
\(\left( 2x-7y+4z-3 \right) +\frac { 1 }{ 6 } \left( 3x-5y+4z+11 \right) =0\)
Multiplying by 6, we get,
\(\Rightarrow 12x-42y+24z-18+3x-5y+4z+11=0\)
\(\Rightarrow 15x-47y+28z-7=0\) which is the required equation of the plane.

Let the intercepts of the plane with the coordinate axes be a, b, c respectively.
∴ Equation of the plane in intercept form is
\(\frac { x }{ a } +\frac { y }{b } +\frac { z }{ c } =1 (1)\)

Given that centroid of ∆ABC is (u, v, w)
\(\therefore (u,v,w)=\left( \frac { a+0+0 }{ 3 } ,\frac { 0+b+0 }{ 3 } ,\frac { 0+0+c }{ 3 } \right) \)
\(\Rightarrow (u,v,w)=\left( \frac { a }{ 3 } ,\frac { b }{ 3 } ,\frac { c }{ 3 } \right) \)
Equating the like co-ordinates we get,
\(\Rightarrow u=\frac { a }{ 3 } \Rightarrow a=3u\)
\(v=\frac { b }{ 3 } \Rightarrow b=3v\)
\(w=\frac { c }{ 3 } \Rightarrow c=3w\)
∴ becomes, 
\(\frac { x }{ 3u } +\frac { y }{ 3v } +\frac { z }{ 3w } =1\)
⇒\(\frac { x }{ u } +\frac { y }{ v } +\frac { z }{ w } =3\)
which is the equation of the required plane.

\(\vec { a } =-\hat { i } +\hat { j } +2\hat { k } \)
Given \(|\vec { n } |=3\sqrt { 3 } \) and let \(\alpha\) be the angle made by the normal with the co-ordinate axes.
\(\therefore { cos }^{ 2 }\alpha +{ cos }^{ 2 }\alpha =1\Rightarrow 3{ cos }^{ 2 }\alpha =1\)
\(\Rightarrow { cos }^{ 2 }\alpha =\frac { 1 }{ 3 } { cos }\alpha =\frac { 1 }{ \sqrt { 3 } } \)
\(\therefore \vec { n } =3\sqrt { 3 } \left( \frac { 1 }{ \sqrt { 3 } } \hat { i } +\frac { 1 }{ \sqrt { 3 } } \hat { j } +\frac { 1 }{ \sqrt { 3 } } \hat { k } \right) =3\hat { i } +3\hat { j } +3\hat { k } \)
∴ The equation of the required plane is
\(\vec { r } .\vec { n } =\vec { a } .\vec { n } \Rightarrow \vec { r } .(3\hat { i } +3\hat { j } +3\hat { k } )\)
\(=(-\hat { i } +\hat { j } +2\hat { k } ).(3\hat { i } +3\hat { j } +3\hat { k } )\)
\(\Rightarrow \vec { r } .(3\hat { i } +3\hat { j } +3\hat { k } )=-3+3+6=6\)
\(\Rightarrow \vec { r } .\frac { (3\hat { i } +3\hat { j } +3\hat { k } ) }{ 2 } =\frac { 6 }{ 3 } =2\)
\(\hat { r } .(\hat { i } +\hat { j } +\hat { k } )\)
Let \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\Rightarrow (x\hat { i } +y\hat { j } +z\hat { k } ).(\hat { i } +\hat { j } +\hat { k } )=2\Rightarrow x+y+z=2\) which is the equation of the required plane.

Given \(\vec { a } \) = \(2\hat { i } +6\hat { j } +3\hat { k } \)
\(\vec { n } \) = \(\hat { i } +3\hat { j } +5\hat { k } \)
Vector form of the evaluation of the plane passing through one point (\(\vec { a } \)) and normal to a vector (\(\vec { n } \)) is
\(\vec { r } .\vec { n } =\vec { a } .\vec { n }\)
\(=\vec { r } .(\hat { i } +3\hat { j } +5\hat { k } )=(2\hat { i } +6\hat { j } +3\hat { k } ).(\hat { i } +3\hat { j } +5\hat { k } )\)
= 2 + 18 + 15
\(\Rightarrow \vec { r } .(\hat { i } +3\hat { j } +5\hat { k } )=35\)
Its Cartesian equation will be
\(a(x-{ x }_{ 1 })+b(y-{ y }_{ 1 })+c(z-{ z }_{ 1 })=0\)
1(x-2)+3(y-6)+5(z-3) = 0
[\(\because ({ x }_{ 1 },{ y }_{ 1 },{ z }_{ 1 }\) is (2, 6, 3) and a, b, c = 1, 3, 5]
\(\Rightarrow\) x-2 + 3y-18 + 5z-15 = 0
\(\Rightarrow\) x + 3y + 5z - 35 = 0
\(\Rightarrow\) x + 3y + 5z = 35

Given cartesian equation of the plane is
12x + 3y - 4z = 65
Its parametric form of vector equation will be
\(\vec { r } .(12\hat { i } +3\hat { j } -4\hat { k } )\) = 65
Here \(\vec { d } =12\hat { i } +3\hat { j } -4\hat { k } \)
\(\frac { 12\hat { i } +3\hat { j } -4\hat { k } }{ \sqrt { { 12 }^{ 2 }+{ 3 }^{ 2 }+(-4)^{ 2 } } } =\frac { 12\hat { i } +3\hat { j } -4\hat { k } }{ \sqrt { 144+9+16 } } \)
\(=\frac { 12\hat { i } +3\hat { j } -4\hat { k } }{ \sqrt { 169 } } \)
\(\hat { d } =\frac { 12\hat { i } +3\hat { j } -4\hat { k } }{ 13 } \)
Hence, the direction cosines of normal to the plane 12x + 3y - 4z = 65 are \(\frac { 12 }{ 13 } ,\frac { 3 }{ 13 } ,\frac { -4 }{ 13 } \)
Also, non-parametric vector form of the equation of the plane is
\(\vec { r } .\left( =\frac { 12\hat { i } +3\hat { j } -4\hat { k } }{ 13 } \right) =\frac { 65 }{ 13 } \)
[From (1)] [Dividing by 13]
\(\Rightarrow \vec { r } .\hat { d } =p\Rightarrow p=\frac { 62 }{ 13 } =5\)

Given lines are
\(\frac { x-1 }{ 2 } =\frac { y+1 }{ 3 } =\frac { z-1 }{ 4 } \)
\(\vec { a } =\hat { i } -\hat { j } +\hat { k } \vec { b } \quad and\quad \vec { b } =2\hat { i } +3\hat { j } +4\hat { k } \)
\(\frac { x-3 }{ 1 } =\frac { y-m }{ 2 } =\frac{z-0}{1}\)\(\Rightarrow \vec { c } =3\hat { i } +m\hat { j } \)
\(\vec { d } =\hat { i } +2\hat { j } +\hat { k } \quad \)
\(\vec { c } -\vec { a } =(3\hat { i } +m\hat { j } )-(\hat { i } -\hat { j } +\hat { k } )\)
\(=2\hat { i } +(m,+1)\hat { j } -\hat { k } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{matrix} \right| =\hat { i } (3-8)-\hat { j } (2-4)+\hat { k } (4-3)\)
\(=-5\hat { i } +2\hat { j } +\hat { k } \)
Since the lines intersect at a point, \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )=0,\)
\(\Rightarrow (2\hat { i } +(m+1)\hat { j } -\hat { k } ).(-5\hat { i } +2\hat { j } +\hat { k } )=0\)
\(\Rightarrow -10+2(m+1)-1=0\Rightarrow -10+2m+2-1=0\)
\(\Rightarrow 2m-9=0\Rightarrow 2m=9\Rightarrow m=\frac { 9 }{ 2 } \)
\(\therefore m=\frac { 9 }{ 2 } \)

Let the given point be \(\vec { a } =5\hat { i } +2\hat { j } +8\hat { k } \)
Given lines are \(\vec { r } =(\hat { i } +\hat { j } -\hat { k } )+s(2\hat { i } -2\hat { j } +\hat { k } )\)
⇒ \(\vec { b } =2\hat { i } -2\hat { j } +\hat { k } \)
and \(\vec { r } =(2\hat { i } -\hat { j } -3\hat { k } )+t(\hat { i } +2\hat { j } +2\hat { k } )\)
⇒ \(\vec { d } =\hat { i } +2\hat { j } +2\hat { k } \)
Since the required line is perpendicular to both \(\vec { b } \) and \(\vec { d } \), it will be parallel to \(\vec { b } \times \vec { d } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & -2 & 1 \\ 1 & 2 & 2 \end{matrix} \right| \)
= \(\hat { i } (-4-2)-\hat { j } (4-1)+\hat { k } (4+2)\)
= \(-6\hat { i } -3\hat { j } +6\hat { k } \)
= \(-3(2\hat { i } +\hat { j } -2\hat { k } )\)
∴ Equation of required straight line is
\(\vec { r } =\vec { a } +m(\vec { b } \times \vec { d } ),m\in R\)
\(\vec { r } =(5\hat { i } +2\hat { j } +8\hat { k } )-3m(2\hat { i } +\hat { j } -2\hat { k } )\)
\(\vec { r } =(5\hat { i } +2\hat { j } +8\hat { k } )+t(2\hat { i } +\hat { j } -2\hat { k } )\)
where t =\(-3m\in R\)
Cartesian equation of a straight line passing through (5, 2, 8) and is 丄 to the straight lines
\(\vec { r } =(\vec { i } +\vec { j } -\vec { k } )+s(2\vec { i } -2\vec { j } +\vec { k } )\)
\(\vec { r } =(2\vec { i } -\vec { j } -3\vec { k } )+t(\vec { i } +2\vec { j } +2\vec { k } )\)
\(\frac { x-{ x }_{ 1 } }{ b_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { b }_{ 2 } } =\frac { z-{ z }_{ 1 } }{ { b }_{ 3 } } \)
\(\frac { x-5 }{ 2 } =\frac { y-2 }{ 1 } =\frac { z-8 }{ -2 } \).

Comparing the given two equations with
\(\vec { r } =\vec { a } +s\vec { b } \) and \(\vec { r } =\vec { c } +s\vec { d } \)
We have \(\vec { a } =2\hat { i } +6\hat {j } +3\hat { k } ,\vec { b } =2\hat { i } +3\hat { j } +4\hat { k } , \vec { c } = 2\hat { j } -3\hat { k } ,\vec { d } =\hat { i } +2\hat { j } +3\hat { k } \)
Clearly, \(\vec { b } \) is not a scalar multiple of \(\vec { d } \). So, the two vectors are not parallel and hence the two lines are not parallel.
The shortest distance between the two straight lines is given by
\(\delta =\frac { \left| (\vec { c } \times \vec { a } ).(\vec { b } \times \vec { d } ) \right| }{ \left| \vec { b } \times \vec { d } \right| } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 4 \\ 1 & 2 & 3 \end{matrix} \right| =\hat { i } -2\hat { j } +\hat { k } \)
\((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )=(-2\hat { i } -4\hat { j } -6\hat { k } ).(\hat { i } -2\hat { j } +\hat { k } )\) = 0
Therefore, the distance between the two given straight lines is zero. Thus, the given lines intersect each other.

The Cartesian equations of the straight line \(\vec { r } =(\hat { i } +\hat { 3j } -\hat { k } )+t(2\hat { i } +3\hat { j } +2\hat { k } )\) is
\(\frac { x-1 }{ 2 } =\frac { y-3 }{ 3 } =\frac { z+1 }{ 2 } \) = s(say)
Then any point on this line is of the form (2s + 1, 3s + 3, 2s -1) ...............(1)
The Cartesian equation of the second line is \(\frac { x-2 }{ 1 } =\frac { y-4 }{ 2 } =\frac { z+3 }{ 4 } =t\) (say)
Then any point on this line is of the form (t + 2, 2t + 4, 4t - 3)....(2)
If the given lines intersect, then there must be a common point. Therefore, for some s, t ∈ R
we have (2s + 1, 3s + 3, 2s −1 ) = (t + 2, 2t + 4, 4t − 3)
Equating the coordinates of x, y and z we get
2s − t = 1, 3s − 2t = 1 and s − 2t =  −1.
Solving the first two of the above three equations, we get s = 1 and t = 1. These values of s and t satisfy the third equation. So, the lines are intersecting.
Now, using the value of s in (1) or the value of t in (2), the point of intersection (3,6,1) of these two straight lines is obtained.
If we take \(\vec{b}=(2\hat { i } +3\hat { j } +2\hat { k } )\) and \(\vec{d}=\hat { i } +\hat { 3j } -\hat { k } \),
then \(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 2 \\ 1 & 2 & 4 \end{matrix} \right| =8\hat { i } -6\hat { j } +\hat { k } \) is a vector perpendicular to both the given straight lines.
Therefore, the required straight line passing through (3,6,1)
and perpendicular to both the given straight lines is the same as the straight line passing through (3,6,1) and parallel to \(8\hat { i } -6\hat { j } +\hat { k } \). Thus, the equation of the required straight line is
\(\vec { r } =(\hat { 3i } +\hat { 6j } -\hat { k } )+m(8\hat { i } -6\hat { j } +\hat { k } )\), k ∈ R.

Every point on the line \(\frac { x-1 }{ 2 } =\frac { y-2 }{ 3 } =\frac { z-3 }{ 4 } \) = s (say) is of the form (2s + 1, 3s + 2, 4s + 3) and every point on the line \(\frac { x-4 }{ 5 } =\frac { y-1 }{ 2 } =z=t\) (say) is of the form (5t + 4, 2t + 1, t).
So, at the point of intersection, for some values of s and t, we have
(2s + 1, 3s + 2, 4s + 3) = (5t + 4, 2t + 1, t)
Therefore, 2s - 5t = 3, 3s - 2t = -1 and 4s - t = -3. Solving the first two equations we get t = -1, s = -1. 
These values of s and t satisfy the third equation. Therefore, the given lines intersect. 
Substituting, these values of t or s in the respective points, the point of intersection is (-1,- 1, -1)

Let the points be A (2, 3, 4), B (-1, 4, 5) and C (8, 1, 2)
Equation of the line joining A and B is
\(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
⇒ \(\frac { x-2 }{ -1-2 } =\frac { y-3 }{ 4- } =\frac { z-4 }{ 5-4 } \)
⇒ \(\frac { x-2 }{ -3 } =\frac { y-3 }{ 1 } =\frac { z-4 }{ 1 } \)
Substitute the point C (8, 1, 2) in line (1),
\(\frac { 8-2 }{ -3 } =\frac { 1-3 }{ 1 } =\frac { 2-4 }{ 1 } \)
⇒ -2 = -2 = -2
Since the point C satisfies the equation of line joining A and B, all the three points lie on the same line.
Hence the given points are collinear.

Given lines are \(\frac { x-5 }{ 5m+2 } =\frac { 2-y }{ 5 } =\frac { 1-z }{ -1 } \)
⇒ \(\frac { x-5 }{ 5m+2 } =\frac { y-2 }{ -5 } =\frac { z-1 }{ 1 } \)
∴ \(\vec { b } =(5m+2)\hat { i } -5\hat { j } +\hat { k } \) ...(1)
and x = \(\frac { 2y+1 }{ 4m } =\frac { 1-z }{ -3 } \)
⇒ \(\frac { x }{ 1 } =\frac { y+\frac { 1 }{ z } }{ 2m } =\frac { z-1 }{ 3 } \)
∴ \(\vec { d } =\hat { i } +2m\hat { j } +3\hat { k } \) ...(2)
since \(\vec { b } \bot \vec { d } \Rightarrow \vec { b } .\vec { d } \)= 0
⇒ \((\hat { i } +2m\hat { j } +3\hat { k } ).((5m+2)\hat { i } -5\hat { j } +\hat { k } )=0\)
⇒ (5m+2)1+2m(-5)+3(1) = 0
⇒ 5m+2-10m+3 = 0
⇒ 5-5m = 0
⇒ 5 = 5m
∴ m = 1

Cartesian equation of straight line passing through two points (2, 1,4) and (a - 1,4, -1) is
\(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
[∵ (x₁, y₁, z₁) = (2, 1, 4) (x₂, y₂, z₂) = (a, -1, 4, -1)]
⇒ \(\frac { x-2 }{ a-1-2 } =\frac { y-1 }{ 4-1 } =\frac { z-4 }{ -1-4 } \)
⇒ \(\frac { x-2 }{ a-3 } =\frac { y-1 }{ 3 } =\frac { z-4 }{ -5 } \)
Similarly, Cartesian equation of straight liens passing through two points (0, 2, b - 1) and (5, 3, -2) is
\(\frac { x-5 }{ 5-0 } =\frac { y-3 }{ 2-3 } =\frac { z+3 }{ b-1+2 } \)
[∵ (x₁,y₁,z₁)=(5,3,-2) (x₂,y₂,z₂) is (0,2,b)...(1)
⇒ \(\frac { x-5 }{ -5 } =\frac { y-3 }{ -1 } =\frac { z+2 }{ b+1 } \)
(1) and (2) are parallel if their direction cosines are equal
∴ Direction ratios ofline (1) are a - 3, 3, -5   ........... (3)
Direction ratios of line (2) are -5, -1, b + 1  ............ (4)
To make the direction ratios equal, multiply (4) by-3.
∴ (4) ⟶ +15, 3, -3b-3
(3) ⟶ a-3, 3, -5
∴ a-3 = +5 ⇒ a = 15+3 = 18
3 = 3
-3b-3 = -15 ⇒ -3b = -5+3 = -2
⇒ b = \(\frac { -2 }{ -3 } =\frac { 2 }{ 3 } \)
∴ a = 18 and b = \(\frac { 2 }{ 3 } \).

The direction ratios of AB is
(6-7, 0- 2, 3 - 1) = (-1, - 2, 2)
[∵ (x₂-x₁), (y₂-y₁), (z₂-z₁)]
Also direction ratios of BC is
(4 - 6, 2 - 0, 4 - 3) = (- 2, 2, 1)
Product of direction ratios is
(- 1) (- 2) + (- 2) (2) + 2 (1)
= 2 - 4 + 2 [∵ if two lines are 丄^r there d.r  d₁b₁ + d₂b₂ + d₃b₃ = 0]
= 4 - 4 = 0
Hence AB 丄 BC
ㄥABC = \(\frac { \pi }{ 2 } \).

Let \(\vec { b } =5\hat { i } +6\hat { j } +7\hat { k } \) and \(\vec { a } =7\hat { i } +9\hat { j } +13\hat { k } \) 
The parametric form of vector equation of a straight line passing through two points \(\vec { a } \) and \(\vec { b } \) is
\(\vec { r } =\vec { a } +t(\vec { b } -\vec { a } )\)
∴ \(\vec { r } =(7\hat { i } +9\hat { j } +13\hat { k } )+t(7-5)\hat { i } +(9-6)\hat { j } +(13-7)\hat { k } \)
\(\vec { r } =(7\hat { i } +9\hat { j } +13\hat { k } )+t(2\hat { i } +3\hat { j } +6\hat { k } ),t\in R\)
The Cartesian equation of a straight line passing through two points as
\(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \) [(x₁,y₁,z₁) is (7,9,13) (x₂,y₂,z₂) is (5,6,7))
\(\frac { x-7 }{ 5-7 } =\frac { y-9 }{ 6-9 } =\frac { z-13 }{ 7-13 } \)
⇒ \(\frac { x-7 }{ -2 } =\frac { y-9 }{ -3 } =\frac { z-13 }{ -6 } \)
= \(\frac { x-7 }{ 2 } =\frac { y-9 }{ 3 } =\frac { z-13 }{ 6 } \)

Let (x₁, y₁, z₁) is (6, 7, 4) and (x₂, y₂, z₂) (8, 4, 9)
The cartesian equation of a straight line passing through two points is
\(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
⇒ \(\frac { x-6 }{ 8-6 } =\frac { y-7 }{ 4-7 } =\frac { z-4 }{ 9-4 } \)
⇒ \(\frac { x-6 }{ 2 } =\frac { y-7 }{ -3 } =\frac { z-4 }{ 5 } \) = s...(1)
⇒ x-6 = 2s 
⇒ y-7 = -3s 
⇒ y = -3s+7
\(\frac { y-7 }{ -3 } \)=s ⇒ y-7 = -3s ⇒ -3s + 7
\(\frac { z-4 }{ 5 } \) = s 
⇒ z-4 = 5s 
⇒ z = 5s + 4
∴ Point on the line is (2s + 6, -3s + 7, 5x + 4)...(2)
To find the point of intersection of (1) and x z plane, put y = 0, in (2)
∴ -3s+7 = 0
⇒ -s = -7 
⇒ s = \(\frac{7}{3}\)
Put s = \(\frac{7}{3}\) in (2) we get, the point of intersection
as \(\left( 2\left( \frac { 7 }{ 3 } \right) +6,0,5\left( \frac { 7 }{ 3 } \right) +4 \right) \)
⇒ \(\left( \frac { 14 }{ 3 } +6,0,\frac { 35 }{ 3 } +4 \right) \)
⇒ \(\left( \frac { 14+18 }{ 3 } ,0,\frac { 35+12 }{ 3 } \right) \Rightarrow \left( \frac { 32 }{ 3 } ,0,\frac { 47 }{ 3 } \right) \)
To find the point of intersection of (1) and yz plane, put x = 0 in  (2)
∴ 2s + 6 = 0
⇒ 2s = -6 
⇒ s = -3
∴ (2) ⟶ (2(-3) + 6, -3(-3) + 7, 5(-3)+4)
= (0, 16, -11)

Let \(\vec { a } =-2\hat { i } +3\hat { j } +4\hat { k } \) and \(\vec { b } =-4\hat { i } +5\hat { j } -6\hat { k } \)
The parametric form of vector equation of a straight line passing through a point \((\vec { b } )\) and parallel to is \(\vec { b } \)is
\(\vec { r } =\vec { a } +t\vec { b } \) where \(t\in R\)
∴ \(\vec { r } =-2\hat { i } +3\hat { j } +4\hat { k } +t(-4\hat { i } +5\hat { j } -6\hat { k } ),t\in R\)
Its Cartesian equation is
\(\frac { x-{ x }_{ 1 } }{ { b }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { b }_{ 2 } } =\frac { z-{ z }_{ 1 } }{ { b }_{ 3 } } \)
⇒ \(\frac { x+2 }{ -4 } =\frac { y-3 }{ 5 } =\frac { z-4 }{ -6 } \)
[∵ (x₁, y₁, z₁) is (-2, 3, 4) & (b₁, b₂, b₃) is (-4, -5, -6)]

Let \(\vec { a } =4\hat { i } +3\hat { j } -7\hat { k } \) and \(\vec { b } =2\hat { i } -6\hat { j } +7\hat { k } \)
Non-parametric form of vector equation of a straight line passing through a point (\(\vec {a} \)) and parallel to a vector (\(\vec { b } \)) is \((\vec { r } -\vec { a } )\times \vec { b } \)
⇒ \([\vec { r } -(4\hat { i } +3\hat { j } -7\hat { k } )]\times (2\hat { i } -6\hat { j } +7\hat { k } )=\vec { 0 } \)
Its cartesian equation is
\(\frac { x-{ x }_{ 1 } }{ { b }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { b }_{ 2 } } =\frac { z-{ z }_{ 1 } }{ { b }_{ 3 } } \)
⇒ \(\frac { x-4 }{ 2 } =\frac { y-3 }{ -6 } =\frac { z+7 }{ 7 } \)
[∵ (x₁, y₁, z₁) is (4, 3, -7) & (b₁, b₂, b₃) is (2, -6, 7)]

Rewriting the given equations as\(​​\frac { x+2 }{ 4 } =\frac { y+3 }{ -2 } =\frac { 2z-6 }{ 3/2 } \) and comparing with \(\frac { x-{ x }_{ 1 } }{ { b }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { b }_{ 2 } } =\frac { z-{ z }_{ 1 } }{ { b }_{ 3 } } \) we have \(\vec { b } ={ b }_{ 1 }\hat { i } +{ b }_{ 2 }\hat { j } +{ b }_{ 3 }\hat { k } \) = \(-4\hat { i } -2\hat { j } +\frac { 3 }{ 2 } \hat { k } =-\frac { 1 }{ 2 } (8\hat { i } +4\hat { j } -3\hat { k } )\). Clearly, \(\vec { b } \) is parallel to the vector \(8\hat { i } +4\hat { j } -3\hat { k } \). Therefore, a vector equation of the required straight line passing through the given point (-4, 2, -3) and parallel to the vector \(8\hat { i } +4\hat { j } -3\hat { k } \) in parametric form is
\(\vec { r } =(-4\hat { i } +2\hat { j } -3\hat { k } )+t(8\hat { i } +4\hat { j } -3\hat { k } )\), t ∈ R
Therefore, Cartesian equations of the required straight line are given by
\(\frac { x-4 }{ 8 } =\frac { y-2 }{ 4 } =\frac { z+3 }{ -3 } \)

\(\vec { a } =\hat { i } +2\hat { j } +3\hat { k } ,\vec { b } =2\hat { i } -\hat { j } +\hat { k } ,\vec { c } =3\hat { i } +2\hat { j } +\hat { k } \)
\(\vec { a } .\vec { c } =(\hat { i } +2\hat { j } +3\hat { k } ),(3\hat { i } +2\hat { j } +\hat { k } )\)
= 3+4+3 = 10
\(\vec { a } .\vec { b } =(\hat { i } +2\hat { j } +3\hat { k } ).(2\hat { i } -\hat { j } +\hat { k } )\)
= 2 - 2 + 3 = 3
∴ \((\vec { a } .\vec { c } )\vec { b } -(\vec { a } .\vec { b } )\vec { c } =10\vec { b } -3\vec { c } \)
⇒ \(\vec { a } \times (\vec { b } \times \vec { c } )=10\vec { b } -3\vec { c } \) ..(1)
\([\vec { a } \times (\vec { b } \times \vec { c } )=(\vec { a } .\vec { c } )\vec { b } -(\vec { a } .\vec { b } )\vec { c } ]\)
Given \(\vec { a } \times (\vec { b } \times \vec { c } )=l\vec { a } +m\vec { b } +n\vec { c } \)..(2)
From (1) & (2)
\(10\vec { b } -3\vec { c } =l\vec { a } +m\vec { b } +n\vec { c } \)
Comparing the co-efficients of like terms, we get
l = 0, m = 10 and n = -3

Given \(\vec { a } ,\vec { b } ,\vec { c } \) and \(\vec { d } \) are co-planar vectors.
\((\vec { a } \times \vec { b } )\times (\vec { c } \times \vec { d } )=[\vec { a } \vec { b } \vec { d } ]\vec { c } -[\vec { a } \vec { b } \vec { c } ]\vec { d } \)...(1)
If \(\vec { a } ,\vec { b } ,\vec { c } \) and \(\vec { d } \) are coplanar vectors then \(\vec { a } ,\vec { b } ,\vec { c } \) are coplanar or \(\vec { a } ,\vec { b } ,\vec { d } \) are also coplanar.
∴ \([\vec { a } \vec { b } \vec { c } ]\) = 0  [∵ They are coplanar]
Also \([\vec { a } \vec { b } \vec { c } ]\)  [∵ they are coplanar]
Substituting these values in (1) we get,
\((\vec { a } \times \vec { b } )\times (\vec { c } \times \vec { d } )=0(\vec { c } )-0(\vec { d } )=\vec { 0 } \)
∴ \((\vec { a } \times \vec { b } )\times (\vec { c } \times \vec { d } )=\vec { 0 } \).

Given \(\vec { a } =-2\hat { i } +5\hat { j } +3\hat { k } \), \(\hat { b } =\hat { i } +3\hat { j } -2\hat { k } \) and \(\vec { c } =-3\vec { i } +\vec { j } +4\vec { k } \)
Volume of the parallelepiped = \(\vec { a } .(\vec { b } \times \vec { c } )\)
= \(\left| \begin{matrix} -2 & 5 & 3 \\ 1 & 3 & -2 \\ -3 & 1 & 4 \end{matrix} \right| \)
= \(-2\left| \begin{matrix} 3 & -2 \\ 1 & 4 \end{matrix} \right| -5\left| \begin{matrix} 1 & -2 \\ - & 4 \end{matrix} \right| +3\left| \begin{matrix} 1 & 3 \\ -3 & 1 \end{matrix} \right| \)
= -2(12+2)-5(4-6)+3(1+9)
= -2(14)-5(-2)+3(10) = -28+10+30
= 12
Vector Area of the base parallelogram = \(\vec { b } \times \vec { c } \)
=\(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 3 & -2 \\ -3 & 1 & 4 \end{matrix} \right| =\hat { i } \left| \begin{matrix} 3 & -2 \\ 1 & 4 \end{matrix} \right| \hat { j } \left| \begin{matrix} 1 & -2 \\ -3 & 4 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 1 & 3 \\ -3 & 1 \end{matrix} \right| \)
= \(\hat { i } \)(12+2)-\(\hat { j } \)(4-6)+\(\hat { k } \)(1+9)
= \(\hat { i } \)(14)-\(\hat { j } \)(-2)+\(\hat { k } \)(10)
= 14\(\hat { k } \)+2\(\hat { j } \)+10\(\hat { k } \)
Area of the parallelogram =\(\sqrt { { 14 }^{ 2 }+{ 2 }^{ 2 }+{ 10 }^{ 2 } } \)
= \(\sqrt { 196+4+100 } \)
= \(\\ \sqrt { 300 } =10\sqrt { 3 } \)..............(2)
Volume of the parallelepiped = Base area x attitude
∴ Altitude = \(\frac { Volume }{ Base\quad area } =\frac { 12 }{ 10\sqrt { 3 } } =\frac { 6 }{ 5\sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } \)

= \(\frac { 2\sqrt { 3 } }{ 5 } \) units.

Given  \(\vec { a } ,\vec { b } ,\vec { c } \) are concurrent edges of a parallelepiped, and its volume is 4 cubic units.
\(\vec { a } .(\vec { b } \times \vec { c } )\) = \(\pm 4\)  ..(1)
Consider
\((\vec { a } +\vec { b } ).(\vec { b } \times \vec { c } )+(\vec { b } +\vec { c } ).(\vec { c } \times \vec { a } )+(\vec { c } +\vec { a } )(\vec { a } \times \vec { b } )\)
= \(\vec { a } .(\vec { b } \times \vec { c } )+\vec { b } .(\vec { b } \times \vec { c } )+\vec { b } .(\vec { c } \times \vec { a } )+\vec { c } .(\vec { c } \times \vec { a } )+\vec { c } .(\vec { a } \times \vec { b } )+\vec { a } .(\vec { a } \times \vec { b } )\)
= \(\vec { a } .(\vec { b } \times \vec { c } )+0+\vec { b } .(\vec { b } \times \vec { c } )+0+\vec { b } .(\vec { c } \times \vec { a } )+0\)
\([\because \vec { a } .(\vec { a } \times \vec { b } )=0]\)
= \(\vec { a } .(\vec { b } \times \vec { c } )+\vec { a } .(\vec { b } \times \vec { c } )+\vec { a } .(\vec { b } \times \vec { c } )\)
\(\left[ \because [\vec { a } \vec { b } \vec { c } ]=[\vec { a } \vec { b } \vec { c } ]=[\vec { a } \vec { b } \vec { c } ] \right] \)
= \(3[\vec { a } .(\vec { b } \times \vec { c } )]=3(\pm 4)\) using (1)
= ±12

Since the vectors \(\vec { a } ,\vec { b } ,\vec { c } \) are coplanar, we have \([\vec { a } ,\vec { b } ,\vec { c } ]\) = 0 Using the properties of the scalar triple product, we get
\([\vec { a } +\vec { b } ,\vec { b } +\vec { c } ,\vec { c } +\vec { a } ]=[\vec { a } ,\vec { b } +\vec { c } ,\vec { c}+\vec {a } ]+[\vec { b } ,\vec { b } +\vec { c } ,\vec { c } +\vec { a } ]\)
\(=[\vec { a } ,\vec { b } ,\vec { c } +\vec { a } ]+[\vec { a } ,\vec { c } ,\vec { c } +\vec { a } ]+[\vec { b } ,\vec { b } ,\vec { c } +\vec { a } ]+[\vec { b } ,\vec { c } ,\vec { c } +\vec { a } ]\)
\(=[\vec { a } ,\vec { b } ,\vec { c } ]+[\vec { a } ,\vec { b } ,\vec { a } ]+[\vec { a } ,\vec { c } ,\vec { c } ]+[\vec { a } ,\vec { c } ,\vec { a } ]+[\vec { b } ,\vec { b } ,\vec { c } ]+[\vec { b } ,\vec { b } ,\vec { a } ]+[\vec { b } ,\vec { c } ,\vec { c } ]+[\vec { b } ,\vec { c } ,\vec { a } ]\)
\(=[\vec { a } ,\vec { b } ,\vec { c } ]+[\vec { a } ,\vec { b } ,\vec { c } ]=2[a,b,c]=0\)
Hence the vectors \(\vec { a } +\vec { b } ,\vec { b } +\vec { c } ,\vec { c } +\vec { a } \) are coplanar.

Resultant force \(\vec { F } =\vec { { F }_{ 1 } } +\vec { F_{ 2 } } +\vec { F_{ 3 } } \)
\(\vec { F } \) = (\(-\hat { 3i } +\hat { 6j } +\hat { 3k } \))+(\(\hat { 4i } -\hat { 10j } +\hat { 12k } \))+(\(\hat { 4i } +\hat { 7j } \))
= \(5\hat { i } +3\hat { j } +9\hat { k } \)
\(\vec { r } \)= (Force acting at the point) - (force acting about the point)
\((8\hat { i } -6\hat { j } -4\hat { k } )-(18\hat { i } +3\hat { j } -9\hat { k } )\)
= \(-10\hat { i } -9\hat { j } +5\hat { k } \)
Torque \((\vec { i } )=\vec { r } \times \vec { F } \)
= \(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ -10 & -9 & 5 \\ 5 & 3 & 9 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} -9 & 5 \\ 3 & 9 \end{matrix} \right| -\hat { j } \left| \begin{matrix} -10 & 5 \\ 5 & 9 \end{matrix} \right| +\hat { k } \left| \begin{matrix} -10 & -9 \\ 5 & 3 \end{matrix} \right| \)
= \(\hat { i } (-18-15)-\hat { j } (-90-25)+\hat { k } (-30+45)\)
\(\vec { i } =-96\hat { i } +115\hat { j } +15\hat { k } \).

Let \(\vec { { F }_{ 1 } } \) and \(\vec { { F }_{ 2 } } \) be the two forces given
Given \(|\vec { { F }_{ 1 } } |=5\sqrt { 2 } \) and its direction is along \(3\hat { i } +4\hat { j } +5\hat { k } \)
∴ \(\vec { { F }_{ 1 } } =5\sqrt { 2 } \) (unit vector of \(3\hat { i } +4\hat { j } +5\hat { k } \))
= \(5\sqrt { 2 } \frac { (3\hat { i } +4\hat { j } +5\hat { k } ) }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 } } } \)   \(\left[ \because \hat { n } =\frac { \vec { n } }{ |\vec { n } | } \right] \)
= \(5\sqrt { 2 } \frac { (3\hat { i } +4\hat { j } +5\hat { k } ) }{ \sqrt { 9+16+25 } } =\frac { 5\sqrt { 2 } (3\hat { i } +4\hat { j } +5\hat { k } ) }{ 5\sqrt { 2 } } \)
= \(3\hat { i } +4\hat { j } +5\hat { k } \)
and  \(\vec { { F }_{ 2 } } =10\sqrt { 2 } \) (unit vector of \(10\hat { i } +6\hat { j } -8\hat { k } \))
= \(10\sqrt { 2 } \frac { (10\hat { i } +6\hat { j } -8\hat { k } ) }{ \sqrt { { 10 }^{ 2 }+{ 6 }^{ 2 }+(-8)^{ 2 } } } \)
\(=\frac{10 \sqrt{\not 2}(10 \hat{i}+6 \hat{j}-8 \hat{k})}{10 \sqrt\not {2}}\)
= \(10\hat { i } +6\hat { j } -8\hat { k } \)
∴ Resistant force \(\vec { F } =\vec { { F }_{ 1 } } +\vec { { F }_{ 2 } } \)
=\((3\hat { i } +4\hat { j } +5\hat { k } )+(10\hat { i } +6\hat { j } -8\hat { k } )\)
\(\vec { F } =13\hat { i } +10\hat { j } -3\hat { k } \)
\(\hat { d } \) = displacement to the point - displacement from the point
= \((6\hat { i } +\hat { j } -3\hat { k } )-(4\hat { i } -3\hat { j } -2\hat { k } )\)
= \(2\hat { i } +4\hat { j } -\hat { k } \)
∴ Work done
w = \(\vec { F } .\vec { d } =(13\hat { i } +10\hat { j } -3\hat { k } ).(2\hat { i } +4\hat { j } -\hat { k } )\)
w = 13(2) + 10(4) - 3(-1) = 26 + 40 + 3
w = 69 units.

Let the forces be \(\vec { { F }_{ 1 } } \) and \(\vec { { F }_{ 2 } } \) and \(\vec { d } \) be the displacement vector
∴ Resultant force \(\vec { F } =\vec { { F }_{ 1 } } +\vec { F_{ 2 } } \)
= \((8\hat { i } +2\hat { j } -6\hat { k } )+(6\hat { i } +2\hat { j } -2\hat { k } )\)
\(\vec { F } =14\hat { i } +4\hat { j } -8\hat { k } \)
\(\vec { d } \)= Displacement to the point - displacement from the point
= \((5\hat { i } +4\hat { j } +\hat { k } )-(\hat { i } +2\hat { j } +\hat { k } )\)
= \(4\hat { i } +2\hat { j } -2\hat { k } \)
Work done (w) = \(\vec { F } .\vec { d } \)
= \((14\hat { i } +4\hat { j } -8\hat { k } ).(4\hat { i } +2\hat { j } -2\hat { k } )\)
= 56 + 8 + 16
= 80 units.

With usual notations in triangle, ABC let \(\vec { BC } =\vec { a } \), \(\vec { CA } =\vec { b } \) and \(\vec { AB } =\vec { c} \). Then \(|\vec { BC }| =\vec { a } \), \(|\vec { CA } |=\vec { b } \) and \(|\vec { AB }| =\vec { c } \)
Since in ΔABC, \(\vec { BC }+\vec { CA }+\vec { AB}=0\) we have \(\vec { BC }\times(\vec { BC }+\vec { CA }+\vec { AB })=\vec { 0 }\)
Simplifying, we get,
\(\vec { BC }\times\vec { CA }=\vec { AB }\times\vec { BC }\) ....(1)

Similarly, since \(\vec { BC }+\vec { CA }+\vec { AB}=\vec 0\), we have
\(\vec {CA} \times (\vec { BC }+\vec { CA }+\vec { AB})=\vec 0\) .... (2)
On Simplification, we obtain \(\vec { BC }\times\vec { CA }=\vec {CA }\times\vec {AB }\)
From equations (1) and (2), we get 
\(\vec { AB }\times\vec {BC }\) = \(\vec { CA }\times\vec {AB }\)=\(\vec { BC}\times\vec {CA}\)
So, \(\left| \overrightarrow { AB} \times \overrightarrow { BC } \right| =\left| \overrightarrow { CA } \times \overrightarrow { AB } \right| =\left| \overrightarrow { BC } \times \overrightarrow { CA } \right| \). Then, we get
ca sin(π − B) = bc sin(π - A) = ab sin (π - C)
That is, ca sin B = bc sin A = absinC . Dividing by abc, we get
\(\frac { sinA }{ A } =\frac { sinB }{ b } =\frac { sinC }{ c } \) or \(\frac { a }{ sinA }= \frac { b }{ sinB } =\frac { c }{ sinC } \)

2x = 3y =  −z \(\Rightarrow \frac{x}{3}=\frac{y}{2}=\frac{-z}{6}\) (Dividing by 6 all)
 \(\Rightarrow \frac{x-0}{3}=\frac{y-0}{2}= \frac{z-0}{-6}\)....(1)
6x = -y = -4z \(\Rightarrow \frac{x}{2}=\frac{-y}{12}=\frac{-z}{3} \) (Dividing by 6 all)
\(\Rightarrow \frac{x-0}{2}=\frac{y-0}{-12}=\frac{z-0}{-3}\) ....(2)
From (1) & (2), we get
\(\vec b = 3\vec i+2\vec j- 6\vec k\)and \( \vec d = 2\vec i-12\vec j- 3\vec k\)
Angle between lines (1) and (2) = Angle between \(\vec b\ and\ \vec d\)
Acute angle between lines cos 0 = \(\frac{|\vec b . \vec d|}{|\vec b||\vec d|} \)
\(\vec b . \vec d \)= (\(\vec b = 3\vec i+2\vec j- 6\vec k\)). (\( 2\vec i-12\vec j- 3\vec k\))
6-24+18 = 0
\( \cos \theta=0 \)
\(\theta=\frac{\pi}{2} \text { or } 90^{\circ}\)

Given lines are \(\frac { x+4 }{ 3 } =\frac { y-7 }{ 4 } =\frac { z+5 }{ 5 } \)
⇒ \(\vec { b } =3\hat { i } +4\hat { j } +5\hat { k } \)
and \(\vec { r } =4\hat { k } +t(2\hat { i } +\hat { j } +\hat { k } )\)
⇒ \(\vec { d } =2\hat { i } +\hat { j } +\hat { k } \)
\( |\vec{b}| =\sqrt{9+16+25}=\sqrt{50} \\ =5 \sqrt{2} \)
∴ cos θ =\(\frac { \vec { b } .\vec { d } }{ |\vec { b } ||\vec { d } | } \)=\(\frac { (3\hat { i } +4\hat { j } +5\hat { k } ).(2\hat { i } +\hat { j } +\hat { k } ) }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 } } .\sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } } } \)
 \( = \frac{15}{5 \sqrt{2} \times \sqrt{6}}=\frac{3}{\sqrt{2} \cdot \sqrt{2} \sqrt{3}}=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2} \\ \theta=\frac{\pi}{6} \)

Length of perpendicular from \(\left( { x }_{ 1 },{ y }_{ 1 },{ z }_{ 1 } \right) \) to the plane.
\(ax+by+cz-p=0\left| \frac { { ax }_{ 1 }+{ by }_{ 1 }+{ cz }_{ 1 }-p }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } \right| \)
\(\therefore \) Length of perpendicular from (1, -2, 3) to the plane
\(x-y-z-5=0\ is\ \delta =\left| \frac { 1-(-2)+3-5 }{ \sqrt { { 1 }^{ 2 }+\left( -1 \right) ^{ 2 }+{ 1 }^{ 2 } } } \right| \)
= \(\left| \frac { 1+2+3-5 }{ \sqrt { 1+1+1 } } \right| =\left| \frac { 1 }{ \sqrt { 3 } } \right| \)
= \(\frac { 1 }{ \sqrt { 3 } } \)

Given lines are \(\vec { r } =(4\hat { i } -\hat { j } )+t(\hat { i } +2\hat { j } -2\hat { k } )\) \([\vec { r } =\vec { a } +t\vec { b } ]\)
∴ \(\vec { b } =\hat { i } +2\hat { j } -2\hat { k } \)
and \(\vec { r } =(\hat { i } -2\hat { j } +4\hat { k } )+s(-\hat { i } -2\hat { j } +2\hat { k } )\)
∴ \(\vec { d } =-\hat { i } -2\hat { j } +2\hat { k } \)
Let θ be the angle between the given lines
Then cos θ = \(\frac { \vec { b } .\vec { d } }{ |\vec { b } ||\vec { d } | } \)
= \(\frac { (\hat { i } +2\hat { j } -2\hat { k } ).(-\hat { i } -2\hat { j } +2\hat { k } ) }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+(-2)^{ 2 } } .\sqrt { (-1)^{ 2 }+{ (-2) }^{ 2 }+{ (2) }^{ 2 } } } \)
= \(\frac { -1-4-4 }{ \sqrt { 9 } .\sqrt { 9 } } =\frac { -9 }{ 9 } \) = -1
∴ cos θ = -1
⇒ cos-1(-1)
⇒ θ = 0

Given \(\vec { a } =\hat { i } +\hat { j } +\hat { k } \), \(\vec { b } =\hat { i } \), \(\vec { c } ={ c }_{ 1 }\hat { i } +{ c }_{ 2 }\hat { j } +{ c }_{ 3 }\hat { k } \)
∴ \(\vec { c } =\hat { i } +2\hat { j } +{ c }_{ 3 }\hat { k } \)
Also, it given that \(\vec { a } ,\vec { b } \) and \(\vec { c } \) are co-planar.
∴ \(\vec { a } .(\vec { b } \times \vec { c } )\)
⇒ \(\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 2 & { c }_{ 3 } \end{matrix} \right| \)= 0
⇒ \(1\left| \begin{matrix} 0 & 0 \\ 2 & { c }_{ 3 } \end{matrix} \right| -1\left| \begin{matrix} 1 & 0 \\ 1 & { c }_{ 3 } \end{matrix} \right| +1\left| \begin{matrix} 1 & 0 \\ 1 & 2 \end{matrix} \right| \) = 0
⇒ 1(0-0)-1(c₃-0)+1(2-0) = 0
⇒ 0 - c₃+2 = 0 ⇒ c₃ = 2
∴ c₃ = 2

Let \(\vec { a } \) = \(2\hat { i } +3\hat { j } +\hat { k } \), \(\vec { b } \)= \(\hat { i } -2\hat { j } +2\hat { k } \) and \(\vec { c } \) = \(\hat { 3i } +\hat { j } +3\hat { k } \)
\(\vec { a } ,\vec { b } \) and \(\vec { c } \) are coplanar if  \(\vec { a } .(\vec { b } \times \vec { c } )\)
Consider \(\vec { a } .(\vec { b } \times \vec { c } )\)
= \(\left| \begin{matrix} 2 & 3 & 1 \\ 1 & -2 & 2 \\ 3 & 1 & 3 \end{matrix} \right| =2\left| \begin{matrix} -2 & 2 \\ 1 & 3 \end{matrix} \right| -3\left| \begin{matrix} 1 & 2 \\ 3 & 3 \end{matrix} \right| +1\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| \)
= 2 (-6- 2) -3 (3 - 6) + 1(1 + 6)
= 2(-8) - 3(-3) + 1(7)
= -16 + 9 + 7
= -16+16
= 0.
Hence, the given vectors are co-planar.