We know that 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 \((\vec { r } .\vec { { n }_{ 1 } } -{ d }_{ 1 })+\lambda (\vec { r } .\vec { { n }_{ 2 } } -{ d }_{ 2 })=0\)
Substituting \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } ,\vec { { n }_{ 1 } } =\hat { i } +\hat { j } +\hat { k } ,\vec { { n }_{ 2 } } =2\hat { i } -3\hat { j } +5\hat { k } \), \({ d }_{ 1 }=1,{ d }_{ 2 }=-2\) in the above equation, we get
(x + y + z + 1) + \(\lambda \) (2x - 3y + 5z - 2) = 0
Since this plane passes through the point (−1, 2,1) , we get λ = \(\frac{3}{5}\), and hence the required equation
of the plane is 11x−4y+20z=1 .

The Cartesian equation of the straight line joining A and B is
\(\frac { x-4 }{ 3 } =\frac { y-1 }{ 4 } =\frac { z-2 }{ 2 } \) = t (say)
Therefore, an arbitrary point on the straight line is of the form (3t + 4, 4t + 1, 2t + 2). 
To find the point of intersection of the straight line and the plane, we substitute x = 3t + 4, y = 4t + 1, z = 2t + 2 in x -y + z = 5 and we get t = 0 Therefore, the point of intersection of the straight line is (4, 1, 2) 
Now, the distance between the two points (4, 1, 2) and (5, -5, -10) is
\(\sqrt { (4-5)^{ 2 }+(1+5)^{ 2 }+(2+10)^{ 2 } } \) = \(\sqrt { 181}\) units.

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
\( (x+2 y+3 z-2)-\frac{7}{2}(x-y+z-3)=0 \)
\(2 x+4 y+6 z-4-7 x+7 y-7 z+21=0 \)
\(-5 x+11 y-z+17=0 \)
\(5 x-11 y+z-17=0\)

\(\frac { x-1 }{ 1 } =\frac { y-2 }{ 1 } =\frac { z-3 }{ { m }^{ 2 } } \) 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,
\(\left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =0 \)
\(\Rightarrow \left( -2\hat { k } \right) .\left[ \left( 4-m^{ 4 } \right) \hat { i } -\hat { j } \left( 2-{ m }^{ 2 } \right) +\hat { k } \left( m^{ 2 }-2 \right) \right] =0\)
\(\Rightarrow 2\left( { m }^{ 2 }-2 \right) =0\)
\(\Rightarrow { m }^{ 2 }-2=0\)
\(\Rightarrow { m }^{ 2 }=2\)
\(\Rightarrow m=\pm \sqrt { 2 } \)

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. 

We observe that the required plane is parallel to the vectors \(\vec { b } =2\hat { i } +3\hat { j } +6\hat { k } \), \(\vec { c } =\hat { i } +\hat { j } -\hat { k } \) and passing through the point(0,1,-5) with position vector \(\vec { a } \). We observe that \(\vec { b } \) is not parallel to \(\vec { c } \).
Then the vector equation of the plane in non-parametric form is given by \((\vec { r } -\vec { a } ).(\vec { b } \times \vec { c } )=0\) ..(1)
Substituting \(\vec { a } =\hat { j } -5\hat { k } \) and \(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 6 \\ 1 & 1 & -1 \end{matrix} \right| =-9\hat { i } +8\hat { j } -\hat { k } \) in equation (1), we get
\((\vec { r } (-\hat { j } -5\hat { k } )).(-9\hat { i } +8\hat { j } -\hat { k } )=0\), which implies that
\(\vec { r } (-9\hat { i } +8\hat { j } -\hat { k } )=13\)
\(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \) 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 -9x + 8y - z = 13 or 9x - 8y + z + 13 = 0

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

Here, \(\vec { a } =(2\hat { i } -\hat { j } +2\hat { k } ),\vec { b } =(3\hat { i } +4\hat { j } +2\hat { k } )\).
The vector form of the given plane is \(\vec { r } .(\hat { i } -\hat { j } +\hat { k } )=5\). Then \(\vec { r } .(\hat { i } -\hat { j } +\hat { k } )=5\) and p = 5
We know that the position vector of the point of intersection of the line  \(\vec { r } =\vec { a } +t\vec { b } \) and the plane
\(\vec { r } .\vec { d } =p\vec { u } =\vec { a } +\left( \frac { p-(\vec { a } .\vec { n } ) }{ \vec { b } .\vec { n } } \right) \vec { b } \), where \(\vec { b } .\vec { n } \neq \vec { 0 } \)
Clearly, we observe that \(\vec { b } .\vec { n } \neq \vec { 0 } \)
Now, \(\frac { p-(\vec { a } .\vec { n } ) }{ \vec { b } .\vec { n } } =\frac { 5-(2\hat { i } -\hat { j } +2\hat { k } ).(\hat { i } -\hat { j } +\hat { k } ) }{ (3\hat { i } +4\hat { j } +2\hat { k } ).(\hat { i } -\hat { j } +\hat { k } ) } =0\). Therefore, the position vector of the point of intersection of the given line and the given plane is
\(\hat { r } =(2\hat { i } -\hat { j } +2\hat { k } )+(0)(3\hat { i } +4\hat { j } +2\hat { k } )=2\hat { i } -\hat { j } +2\hat { k } \)
That is, the given straight line intersects the plane at the point (2, −1, 2)
Aliter:
The Cartesian equation of the given straight line is \(\frac { x-2 }{ 3 } =\frac { y+1 }{ 4 } =\frac { z-2 }{ 2 } =t\)(say)
We know that any point on the given straight line is of the form (3t+2, 4 t−1, 2 t+2). If the given line and the plane intersects, then this point lies on the given pane x−y+z−5 = 0.
So, (3t + 2)−(4t − 1) + (2t + 2) − 5 = 0 ⇒ t = 0.
Therefore, the given line intersects the given plane at the point (2, -1, 2)

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.