From the line \(\frac { x-1 }{ 3 } =\frac { y+1 }{ 2 } =\frac { z-1 }{ 5 } \)
(x₁, y₁, z₁) is (1, -1, 1)
(l₁, m₁, n₁) is (3, 2, 5)
From the line \(\frac { x+2 }{ 4 } =\frac { y-1 }{ 3 } =\frac { z+1 }{ -2 } \)
we get, (x₂, y₂, z₂) is (-2, 1, -1)
(l₂, m₂, n₂) is 4, 3, -2
The Condition for intersecting lines is 
\(\left| \begin{matrix} { x }_{ 2 }-{ x }_{ 1 } \\ { l }_{ 1 } \\ { l }_{ 2 } \end{matrix}\begin{matrix} { y }_{ 2 }-{ y }_{ 1 } \\ { m }_{ 1 } \\ { m }_{ 2 } \end{matrix}\begin{matrix} { z }_{ 2 }-{ z }_{ 1 } \\ { n }_{ 1 } \\ { n }_{ 2 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} -2-1 \\ 3 \\ 4 \end{matrix}\begin{matrix} 1+1 \\ 2 \\ 3 \end{matrix}\begin{matrix} -1-1 \\ 5 \\ -2 \end{matrix} \right| \)
\(\Rightarrow \left| \begin{matrix} -3 \\ 3 \\ 4 \end{matrix}\begin{matrix} 2 \\ 2 \\ 3 \end{matrix}\begin{matrix} -2 \\ 5 \\ -2 \end{matrix} \right| \)
= -3 (-4 -15) -2 (-6 -20) -2 (9 - 8)
= -3(-19) - 2(-26) -2 (1)
= 57 + 52 - 2 = 57 + 50
= 107 ≠ 0
Hence the given lines do not intersect

L. H. S = \(\left[ \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ c } \right] \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \times \overset { \rightarrow }{ c } \right\} \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ c } \right\} \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right\} \ \left[ \because \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ c } =\overset { \rightarrow }{ 0 } \right] \)
\(=\overset { \rightarrow }{ a } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) +\overset { \rightarrow }{ b } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) +\overset { \rightarrow }{ c } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \)
\(=\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] +0+0\)
\(\left[ \because \left[ \overset { \rightarrow }{ b } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =\left[ \overset { \rightarrow }{ c } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =0 \right] \)
\(=\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] \)= R. H. S
Hence proved

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]
\(\Rightarrow \vec { r } .\hat { d } =p\Rightarrow p=\frac { 62 }{ 13 } =5\)

The equation of the plane having intercepts a, b, c on the x, y, z axes respectively is \(\frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =1\). 
Since the sum of the reciprocals of the intercepts on the coordinate axes is a constant, we have \(\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } =k\),  where k is a constant, and which can be written as \(\frac { 1 }{ a } \left( \frac { 1 }{ k } \right) +\frac { 1 }{ b } \left( \frac { 1 }{ k } \right) +\frac { 1 }{ c } \left( \frac { 1 }{ k } \right) =1\)
This shows that the plane \(\frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =1\) passes through the fixed point \(\left( \frac { 1 }{ k } ,\frac { 1 }{ k } ,\frac { 1 }{ k } \right) \)

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.

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.

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)]

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
\(\frac { x-4 }{ 8 } =\frac { y-2 }{ 4 } =\frac { z+3 }{ -3 } \)

Treating \((\vec { a } \times \vec { b } )\) as the first vector on the right hand side of the given equation and using the vector triple product expansion, we get
\((\vec { a } \times \vec { b } )\times (\vec { a } \times \vec { c } )=((\vec { a } \times \vec { b } ).\vec { c } )\vec { a } -((\vec { a } \times \vec { b } ).\vec { a } )\vec { c } =(\vec { a } .\vec { b } \times \vec { c } ))\vec { a } \)

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 position vectors of the parts A and B on the circle lie \(\vec { a } \) and \(\vec { b } \) respectively.
Since O is the centre of the circle
\(|\vec { OA } |=|\vec { OB } |\Rightarrow |\vec { a } |=|\vec { b } |\) ....(1)
Also D is the mid-point of AB,
⇒ \(\vec { OD } =\frac { \vec { a } +\vec { b } }{ 2 } \) (mid-point formula)
\(\left( \frac { \vec { a } +\vec { b } }{ 2 } \right) .(\vec { OB } -\vec { OA } )\)
=\(\left( \frac { \vec { a } +\vec { b } }{ 2 } \right) .(\vec { Ob } -\vec { Oa } )\)
= \(\frac { 1 }{ 2 } \left[ |\vec { b } |^{ 2 }-|\vec { a}| ^{ 2 } \right] \)
=\(\left[ \because (\vec { a } +\vec { b } ).(\vec { b } -\vec { a } )=|\vec { b } |^{ 2 }-|\vec { a } |^{ 2 } \right] \)
= \(\frac { 1 }{ 2 } \left[ |\vec { b } |^{ 2 }-|\vec { b| } ^{ 2 } \right] \) (using (1))
= \(\frac{1}{2}\)(0) = 0
⇒ \(\vec { OD } .\vec { AB } =0\Rightarrow \vec { OD } \bot \vec { AB } \)
Hence, if a line is drawn from the centre of a to the mid-point of a chord, then that line is perpendicular to the chord.

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 \(\frac { a }{ sinA }= \frac { b }{ sinB } =\frac { c }{ sinC } \)

With usual notations in triangle ABC, we have \(\vec { BC } =\vec { a } ,\vec { CA } =\vec { b } \) \(\vec { AB } =\vec { c } \).
Then \(\left| \vec { BC } \right| =a\) , \(\left| \vec { CA } \right| =b\), \(\left| \vec { AB} \right| =c\), and \(\vec { BC } +\vec { CA } +\vec { AB } =\vec { 0 } \)
So, \(\vec { BC } =-\vec { CA } -\vec { AB } \)
Then applying dot product, we get
\(\vec { BC } .\vec { BC } =(-\vec { CA } -\vec { AB } ).(-\vec { CA } -\vec { AB } )\)
⇒ \({ \left| \vec { BC } \right| }^{ 2 }={ \left| \vec { CA } \right| }^{ 2 }+{ \left| \vec { AB } \right| }^{ 2 }+\vec { 2CA } .\vec { AB } \)
⇒ a² = b²+c²+2bc cos (\(\pi\) - A)
⇒ a² = b²+c²−2bc cos A.
The results (ii) and (iii) are proved in a similar way.

Given \(\overset { \rightarrow }{ OA } =2\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } ,\overset { \rightarrow }{ OB } =\overset { \wedge }{ i } -3\overset { \wedge }{ j } -5\overset { \wedge }{ k } ,\) and \(\overset { \rightarrow }{ OC } =3\overset { \wedge }{ i } -4\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)

\(\therefore \overset { \rightarrow }{ AB } =\overset { \rightarrow }{ OB } -\overset { \rightarrow }{ OA } =-\overset { \wedge }{ i } -2\overset { \wedge }{ j } -6\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ BC } =\overset { \rightarrow }{ OC } -\overset { \rightarrow }{ OB } =2\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ CA } =\overset { \rightarrow }{ OA } -\overset { \rightarrow }{ OC } =\overset { \wedge }{ -i } +3\overset { \wedge }{ j } +5\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ AB } +\overset { \rightarrow }{ BC } +\overset { \rightarrow }{ CA } =\overset { \rightarrow }{ 0 } \)
Also, \(\overset { \rightarrow }{ BC } \). \(\overset { \rightarrow }{ CA } \)=\(\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ -i } +3\overset { \wedge }{ j } +5\overset { \wedge }{ k } \right) \)
= -2 -3 + 5 = 0
\(\Rightarrow \overset { \rightarrow }{ BC } .\overset { \rightarrow }{ CA } \Rightarrow \angle BCA=\frac { \pi }{ 2 } \)
Hence, ABC is a right angled triangle.
\(\cos { A } =\frac { \overset { \rightarrow }{ AB } .\overset { \rightarrow }{ AC } }{ \left| \overset { \rightarrow }{ AB } \right| \left| \overset { \rightarrow }{ AC } \right| } \)
\(=\frac { \left( -\overset { \wedge }{ i } -2\overset { \wedge }{ j } -6\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ -i } -3\overset { \wedge }{ j } -5\overset { \wedge }{ k } \right) }{ \sqrt { 1+4+36 } .\sqrt { 1+9+25 } } \)
\(=\frac { 35 }{ \sqrt { 41 } .\sqrt { 35 } } =\sqrt { \frac { 35 }{ 41 } } =A={ Cos }^{ -1 }\left( \sqrt { \frac { 35 }{ 41 } } \right) \)
\(\cos { B } =\frac { \overset { \rightarrow }{ BA. } \overset { \rightarrow }{ BC } }{ \left| \overset { \rightarrow }{ BA } \right| \left| \overset { \rightarrow }{ BC } \right| } =\frac { \left( \overset { \wedge }{ i } +2\overset { \wedge }{ j } +6\overset { \wedge }{ k } \right) .\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) }{ \sqrt { 1+4+36 } +\sqrt { 4+1+1 } } \)
\(=\sqrt { \frac { 6 }{ 41 } } \Rightarrow B={ Cos }^{ -1 }\left( \sqrt { \frac { 6 }{ 41 } } \right) \)

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

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 axis
Then \(|\vec { OL } |=|\vec { OA } |\) = cos α = cos α
\(\\ |\vec { LA } |=|\vec { OB } |\) sin α = sin α
[∵ ΔOAL, sin α = \(\frac { opp }{ hyp } cos\alpha =\frac { adj }{ hyp } \)]
∴ \(\vec { OL } =|\vec { OL } |\hat { i } =cos\alpha \hat { i } \)
\(\vec { LA } =|\vec { LA } |\hat { j } =sin\ \alpha \hat { j } \)
∴ \(\hat { a } =\vec { OA } =\vec { OL } +\vec { LA } \) [Using Δ law of addition]
\(\hat { a } =cos\alpha \hat { i } +sin\alpha \hat { j } \)
Similarly \(\\ \hat { b } =cos\alpha \hat { i } +sin\alpha \hat { j } \) ...(2)
The angle between \(\hat { a } \) and \(\hat { b } \) is (α-β)
∴ \(\hat { a } .\hat { b } =|\vec { a } ||\vec { b } |\)cos(α-β) = cos(α-β)..(3)
[∵ \(|\vec { a } |=|b|\) = 1]
Also \(\vec { a }. \vec { b } =(cos\alpha \hat { i } +sin\alpha \hat { j } )(cos\beta \hat { i } +sin\beta \hat { j } )\)
\(\hat { a } \hat { b } \) = cosα cosβ + sinα sinβ....(4)
From (3) and (4), we get
cos(α-β) = cos α cos β + sin α sin β

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

Given (x₁, y₁, z₁) is (2, -1, 3) (x₂, y₂, z₂) is (4, 2, 1)
Cartesian equation of a line passing through two points is \(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
\(\Rightarrow \frac { x-2 }{ 2 } =\frac { y+1 }{ 3 } =\frac { z-3 }{ -2 } \)

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.

Let \(\vec { a } =-6\hat { i } +14\hat { j } +10\hat { k } \), \(\vec { b } =14\hat { i } -10\hat { j } -6\hat { k } \) and \(\vec { c } =2\hat { i } +4\hat { j } -2\hat { k } \)
Volume of the parallelepiped having \(\vec { a } ,\vec { b } \) and \(\vec { c } \) as its co-terminus edges is \(\vec { a } .(\vec { b } \times \vec { c } )\).
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} -6 & 14 & 10 \\ 14 & -10 & -6 \\ 2 & 4 & -2 \end{matrix} \right| \)
= \(-6\left| \begin{matrix} -10 & -6 \\ 4 & -2 \end{matrix} \right| -14\left| \begin{matrix} 14 & -6 \\ 2 & -2 \end{matrix} \right| +10\left| \begin{matrix} 14 & -10 \\ 2 & 4 \end{matrix} \right| \)
= -6(20 + 24) - 14(-28 + 12) + 10(56 + 20)
= -6(44) -14(-16) + 10(76)
= -264 + 224 + 760 = 720.
∴ Volume of the required parallelepiped = 720 cubic units.

Since the given three vectors are coplanar, we have \(\left| \begin{matrix} 2 & -1 & 3 \\ 3 & 2 & 1 \\ 1 & m & 4 \end{matrix} \right| \) = 0 ⇒ m = -3

By the defination of scalar triple product of three vectors,
We find, \(\hat { a } .(\hat { b } \times \hat { c } )\) = \(\left| \begin{matrix} -3 & -1 & 5 \\ 1 & -2 & 1 \\ 0 & 4 & -5 \end{matrix} \right| =-3\)