\(\overset { \rightarrow }{ OA } =3\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \) , \(\overset { \rightarrow }{ OB } =\overset { \wedge }{ i } -\overset { \wedge }{ j } -3\overset { \wedge }{ k } \), \(\overset { \rightarrow }{ OC } =4\overset { \wedge }{ i } -3\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
Area of △ ABC = \(\frac { 1 }{ 2 } \left| \overset { \rightarrow }{ AB } \times \overset { \rightarrow }{ AC } \right| \)
\(\overset { \rightarrow }{ AB } =\overset { \rightarrow }{ OB } -\overset { \rightarrow }{ OA } =\left( \overset { \wedge }{ i } -\overset { \wedge }{ j } -3\overset { \wedge }{ k } \right) -\left( 3\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) =-2\overset { \wedge }{ i } -5\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ AC } =\overset { \rightarrow }{ OC } -\overset { \rightarrow }{ OA } =\left( 4\overset { \wedge }{ i } -3\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) -\left( 3\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) =\overset { \wedge }{ i } -2\overset { \wedge }{ j } -\overset { \wedge }{ k } \)
\(\therefore \overset { \rightarrow }{ AB } \overset { \rightarrow }{ \times AC } =\left| \begin{matrix} \overset { \wedge }{ i } \\ -2 \\ 1 \end{matrix}\begin{matrix} \overset { \wedge }{ j } \\ 0 \\ -2 \end{matrix}\begin{matrix} \overset { \wedge }{ k } \\ -5 \\ -1 \end{matrix} \right| \)
\(=\overset { \wedge }{ i } \)(0-10) - \(\overset { \wedge }{ j } \) (2+5) + \(\overset { \wedge }{ k } \) (4-0)
\(=10\overset { \wedge }{ i } -7\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
\(\therefore \left| \overset { \rightarrow }{ AB } \overset { \rightarrow }{ \times AC } \right| =\sqrt { 101+49+16 } =\sqrt { 165 } \)
∴ Area of Δ ABC \(=\frac { 1 }{ 2 } \sqrt { 165 } \) sq. units 

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

Let \(\vec { a } ={ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } \)
∴ LHS = \(\hat { i } \times (\vec { a } \times \hat { i } )+\hat { j } \times (\vec { a } \times \hat { j } )+\hat { k } \times \vec { a } \times \hat { k }\)
\((\hat { i }. \hat { i } )\vec { a } -(\hat { i } .\hat { a } )\hat { i } +(\hat { j } .\hat { j } )\vec { a } -(\hat { j } .\vec { a } )\hat { j } +(\hat { k } .\hat { k } )\vec { a } -(\hat { k } .\vec { a } )\hat { k } \)
\([\because \vec { a } \times (\vec { b } \times \vec { c } )=(\vec { a } .\vec { c } )\vec { b } -(\vec { a } .\vec { b } )\vec { c } ]\)
\(1.\vec { a } -{ a }_{ 1 }\hat { i } +1.\vec { a } -{ a }_{ 2 }\hat { j } +1.\vec { a } -{ a }_{ 3 }\hat { k } ]\)
\([\because \hat { i } .\hat { i } =\hat { j } .\hat { j } =\hat { k } .\hat { k } =1\)and
\(\hat { i } \vec { a } =\hat { i } ({ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } )={ a }_{ 1 }\hat { j } .\vec { a } ={ a }_{ 2 }\quad \hat { k } .\vec { a } ={ a }_{ 3 }\)
\(3\vec { a } -({ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } )\)
= \(3\vec { a } -\vec { a } =2\vec { a } \)
= RHS .
∴ LHS = RHS. Hence proved

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

\(\vec{p} \times \vec{q}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -3 & 4 & -7 \\ 6 & 2 & -3 \end{array}\right|=2 \hat{i}-51 \hat{j}-30 \hat{k}\)
Now, \(\vec{p} \cdot(\vec{p} \times \vec{q})=(-3 \hat{i}+4 \hat{j}-7 \hat{k}) \cdot(2 \hat{i}-51 \hat{j}-30 \hat{k})\)
= -6 - 204 + 210 = 0
Hence \(\vec q\) and \(\vec p \times \vec q\)are pefpendicular to each other.
Now, \(\vec{q} \cdot(\vec{p} \times \vec{q})=(-6 \hat{i}+2 \hat{j}-3 \hat{k}) \cdot(2 \hat{i}-51 \hat{j}-30 \hat{k})\)
= 12 - 102 + 90 = 0
Hence \(\vec q\) and \(\vec p \times \vec q\) are perpendicular to each other.

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)

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 \(\frac { x-1 }{ 4 } =\frac { y-2 }{ 5 } =\frac { z+3 }{ -7 } \)

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 A be the point (0, -2, 3)
\( \overrightarrow{\mathrm{r}} =\overrightarrow{\mathrm{OA}}=-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \)
\(\overrightarrow{\mathrm{F}} =\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \)
\(\overrightarrow{\mathrm{t}} =\overrightarrow{\mathrm{r}} \times \vec{F}=\left|\begin{array}{lll} \hat{\mathrm{i}} \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 0 & 2 & 3 \\ 1 & 1 & 1 \end{array}\right| \)
\( =\hat{\mathrm{i}}(-2-3)-\hat{\mathrm{j}}(0-3)+\hat{\mathrm{k}}(0+2) \)
\(\overrightarrow{\mathrm{t}} =-5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
Magnitude of moment \(=\sqrt{(25)+(9)+(4)}=\sqrt{38}\)
Direction cosines \(=\left(\frac{-5}{\sqrt{38}}, \frac{3}{\sqrt{38}}, \frac{2}{\sqrt{38}}\right)\)

Comparing the given lines with the general Cartesian equations of straight lines,
\(\frac { x-{ x }_{ 1 } }{ { b }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { b }_{ 2 } } =\frac { z-{ z }_{ 1 } }{ { b }_{ 3 } } \)and \(\frac { x-{ x }_{ 2 } }{ { d }_{ 1 } } =\frac { y-{ y }_{ 2 } }{ { d }_{ 2 } } =\frac { z-{ z }_{ 2 } }{ { d }_{ 3 } } \)
we find (b₁, b₂, b₃) = (2, 1, -2) and (d₁, d₂, d₃) = (4, -4, 2). Therefore, the angle between the two straight lines is 
\(\theta ={ cos }^{ -1 }\left( \frac { \left| (2)(4)+(1)(-4)+(-2)(2) \right| }{ \sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 }+({ -2) }^{ 2 } } \sqrt { { 4 }^{ 2 }+(-{ 4 })^{ 2 }+{ 2 }^{ 2 } } } \right) ={ cos }^{ -1 }\left( 0 \right) =\frac { \pi }{ 2 } \)
Thus the two straight lines are perpendicular.

If \(\hat { b } =\frac { 2\hat { i } +2\hat { j } -\hat { k } }{ \left| \hat { 2i } +2\hat { j } -\hat { k } \right| } =\frac { 1 }{ 3 } (2\hat { i } +2\hat { j } -\hat { k } )\). Therefore from the definition of direction cosines of \(\hat { b } \), we have
\(cos\alpha =\frac { 2 }{ 3 } ,cos\beta =\frac { 2 }{ 3 } ,cos\gamma =-\frac { 1 }{ 3 } \)
where α, β, \(\gamma \) are the angles made by \(\hat { b } \) with the positive x -axis, positive y -axis, and positive
z -axis, respectively. As the angle between the given straight line with the coordinate axes are same as the angles made by \(\hat { b } \) with the coordinate axes, we have \(\alpha ={ cos }^{ -1 }\left( \frac { 2 }{ 3 } \right) ,\beta ={ cos }^{ -1 }\left( \frac { 2 }{ 3 } \right) ,\gamma ={ cos }^{ -1 }\left( \frac { -1 }{ 3 } \right) \) respectively.

Given  \(\hat { a } \times (\hat { b } \times \hat { c } )=\frac { 1 }{ 2 } \hat { b } \)
⇒ \((\hat { a } .\hat { c } )\hat { b } -(\hat { a } .\hat { b } )\hat { c } =\frac { 1 }{ 2 } \hat { b } \)
⇒ \(\lambda \hat { b } -\mu \hat { c } =\frac { 1 }{ 2 } \hat { b } \) [∵ put \(\lambda =\hat { a } .\hat { c } \) & \(\mu =\hat { a } .\hat { b } \)]
⇒ \(\left( \lambda -\frac { 1 }{ 2 } \right) \hat { b } -\mu \hat { c } \) = 0
Since \(\hat { b } \) and \(\hat { c } \) are non-collinear vectors
\(\lambda -\frac { 1 }{ 2 } \) = 0 and μ = 0
∴ \(\lambda =\frac { 1 }{ 2 } \)
⇒ \(\hat { a } .\hat { c } =\frac { 1 }{ 2 } \)
⇒ \(|\hat { a } ||\hat { c } |cos\theta =\frac { 1 }{ 2 } \)  [∵ By the definition of scalar product]
⇒ (1) (1) \(cos\theta =\frac { 1 }{ 2 } \)    \(\left[ \because |\vec { a } |=|\vec { c } |=1 \right] \)
⇒ cos θ = \(\frac{1}{2}\)
⇒ θ = 60⁰ = \(\frac { \pi }{ 3 } \)
Hence angle between \(\vec { a } \) and \(\vec { c } \) is \(\frac { \pi }{ 3 } \).

Using the definition of the scalar triple product, we get 
\([\vec { a } \times \vec { b } ,\vec { b } \times \vec { c } ,\vec { c } \times \vec { a } ]\) = \((\vec { a } \times \vec { b } ).[(\vec { b } \times \vec { c } )\times (\vec { c } \times \vec { a } )]\) ....(1)
By treating \((\vec { b } \times \vec { c } )\) as the first vector in the vector triple product, we find
\((\vec { b } \times \vec { c } )\times (\vec { c } \times \vec { a } )\) = \(((\vec { b } \times \vec { c } ).\vec { a } )\vec { c } \) -  \(((\vec { b } \times \vec { c } ).\vec { c } )\vec { a } )\) = \([{ \vec { a } ,\vec { b } ,\vec { c } }]\vec { c } \)
Using this value in (1), we get
\([\vec { a } \times \vec { b } ,\vec { b } \times \vec { c } ,\vec { c } \times \vec { a } ]\) = \((\vec { a } \times \vec { b } ).([\vec { a } ,\vec { b } ,\vec { c } ]\vec { c } )=[\vec { a } ,\vec { b } ,\vec { c } ](\vec { a } \times \vec { b } ).\vec { c } ={ [\vec { a } ,\vec { b } ,\vec { c } ] }^{ 2 }\)

  \(|\vec { c } |\) = 1 and \(\vec { c } \bot \vec { a } \) & \(\vec { b } \)
Also, angle between \(\vec { a } \) and \(\vec { b } \) is \(\frac { \pi }{ 6 } \)
Consider \([\vec { a } \vec { b } \vec { c } ]=\vec { a } .(\vec { b } \times \vec { c } )\)
= \((\vec { a } \times \vec { b } ).\vec { c } \)
[∵ angle between \(\vec { a } \) and \(\vec { b } \) is \(\frac { \pi }{ 6 } \). \(\vec { c } \) 丄 both a & b]
= \(|\vec { a } ||\vec { b } |sin\frac { \pi }{ 6 } .\vec { c } .\vec { c } \)
= \(|\vec { a } ||\vec { b } |.\frac { 1 }{ 2 } \)(1)
= \(|\vec { a } ||\vec { b } |.\frac { 1 }{ 2 } \) [∵ \(\vec { c } .\vec { c } \) = 1]
∴ \([\vec { a } \vec { b } \vec { c } ]^{ 2 }=|\vec { a } |^{ 2 }|\vec { b } |^{ 2 }.\frac { 1 }{ 4 } =\frac { 1 }{ 4 } |\vec { a } |^{ 2 }|\vec { b } |^{ 2 }\).

We get \([\vec { a } +\vec { c } ,\vec { a } +\vec { b } ,\vec { a } +\vec { b } +\vec { c } ]\) = \(\left| \begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{matrix} \right| \)\([\vec { a } ,\vec { b } ,\vec { c } ]\) = \([\vec { a } ,\vec { b } ,\vec { c } ]\)