We have 
\( \vec{F} =3 \hat{i}+2 \hat{j}-4 \hat{k} \)
\(\overrightarrow{O P} =\hat{i}-\hat{j}+2 \hat{k} \)
\(\overrightarrow{O A} =2 \hat{i}-\hat{j}+3 \hat{k} \)
\(\vec{r} =\overrightarrow{A P}=\overrightarrow{O P}-\overrightarrow{O A} \)
\( =(\hat{i}-\hat{j}+2 \hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k}) \)
\(\vec{r} =-\hat{i}-\hat{k}\)
The moment \(\vec M\) of the force \(\vec F\) about the point A is given by
\(\vec{M}=\vec{r} \times \vec{F}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 0 & -1 \\ 3 & 2 & -4 \end{array}\right|=2 \hat{i}-7 \hat{j}-2 \hat{k}\)

Let \(\vec{a}=2 \hat{i}+\hat{j}-\hat{k} ; \vec{b}=\hat{i}+2 \hat{j}+\hat{k}\)
Let \(\theta\) be the angle between \(\vec a\ and\ \vec b\) 
\( \therefore \theta =\sin ^{-1}\left(\frac{\mid \vec{a} \times \vec{b}}{|\vec{a}| \mid \vec{b}} \mid\right) \)
\(\vec{a} \times \vec{b} =\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 2 & 1 \end{array}\right|=3 \hat{i}-3 \hat{j}+3 \hat{k} \)
\(|\vec{a} \times \vec{b}| =\sqrt{3^{2}+(-3)^{2}+3^{2}}=3 \sqrt{3} \)
\(|\vec{a}| =\sqrt{2^{2}+1^{2}+(-1)^{2}}=\sqrt{6} \)
\(|\vec{b}| =\sqrt{1^{2}+2^{2}+1^{2}}=\sqrt{6} \)
\(\therefore \theta =\sin ^{-1}\left(\frac{3 \sqrt{3}}{\sqrt{6} \sqrt{6}}\right) \)
\( =\sin ^{-1}\left(\frac{3 \sqrt{3}}{6}\right) \)
\( =\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right) \)
\(\therefore \theta =\frac{\pi}{3} \)

\(\vec{F}=a \hat{i}+\hat{j}+\hat{k} ; \overrightarrow{O A}=\hat{i}+\hat{j}+\hat{k} ; \overrightarrow{O B}=2 \hat{i}+2 \hat{j}+2 \hat{k}\)
Work done = 5 units
\(\vec{d}=\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\hat{i}+\hat{j}+\hat{k}\)
Work done = \(\vec{F} \cdot \vec{d}\)
\( 5 =(a \hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k}) \)
\(5 =a+1+1 \Rightarrow a=3 \)
\(|\vec{a} \times \vec{b}| =\sqrt{(-1)^{2}+(2)^{2}+(2)^{2}}=3 \)
Required vectors \( =6\left[\pm\left[\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\right)\right] \)
\( =\pm(-2 \hat{i}+4 \hat{j}+4 \hat{k})\)

\( \vec{F}=\hat{i}+3 \hat{j}-\hat{k} ; \overrightarrow{O A} =2 \hat{i}-6 \hat{j}+7 \hat{k} ; \overrightarrow{O B}=3 \hat{i}-\hat{j}-5 \hat{k} \)
\(\vec{d} =\overrightarrow{A B} \)
\( =\overrightarrow{O B}-\overrightarrow{O A}=\hat{i}+5 \hat{j}-12 \hat{k} \)
Work done \( =\vec{F} \cdot \vec{d} \)
\( =(\hat{i}+3 \hat{j}-\hat{k}) \cdot(\hat{i}+5 \hat{j}-12 \hat{k}) \)
\( =(1)(1)+3(5)+12=28 \)

Let \(\vec{a}=4 \hat{i}-\hat{j}+3 \hat{k} ; \quad \vec{b}=-2 \hat{i}+\hat{j}-2 \hat{k}\)
\( \vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{array}\right|=-\hat{i}+2 \hat{j}+2 \hat{k} \)
\( |\vec{a} \times \vec{b}|=\sqrt{1+4+4}=\sqrt{9}=3 \)
\( \hat{\mathrm{n}}=\pm \frac{6(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|} \)
\( =\pm \frac{^2\not6(-\hat{i}+2 \hat{j}+2 \hat{k})}{\not 3}=\pm 2(-\hat{i}+2 \hat{j}+2 \hat{k}) \)

\(\vec{p} \times \vec{q}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -3 & 4 & -7 \\ 6 & 2 & -3 \end{array}\right|=2 \hat{i}-51 \hat{j}-30 \hat{k}\)
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.

\(\overrightarrow{O A}=\hat{i}+2 \hat{j}+3 \hat{k}; \overrightarrow{O B}=4 \hat{i}+\hat{j}+5 \hat{k} ; \overrightarrow{O C}=7 \hat{i}+7 \hat{k}\)
\( \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=(4 \hat{i}+\hat{j}+5 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k}) \)
\(\overrightarrow{A B}=3 \hat{i}-\hat{j}+2 \hat{k} \)
\(\overrightarrow{A C}=\overrightarrow{O C}-\overrightarrow{O A}=6 \hat{i}-2 \hat{j}+4 \hat{k} \)
\(\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc} i & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 6 & -2 & 4 \end{array}\right|=\overrightarrow{0} \)
The vectors \(\overline{AB}\) and \(\overline {AC}\) are parallel. But they have the point A as a common point.
\(\overline{AB}\) and \(\overline {AC}\) are along same line
\(\therefore\)  A, B, C are collinear.

\( {[\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}]} \)
\( =[\vec{a}, \vec{b}+\vec{c}, \vec{c}+\vec{a}]+[\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}] \)
\( =[\vec{a}, \vec{b}, \vec{c}+\vec{a}]+[\vec{a}, \vec{c}, \vec{c}+\vec{a}]+[\vec{b}, \vec{b}, \vec{c}+\vec{a}]+[\vec{b}, \vec{c}, \vec{c}+\vec{a}] \)
\(=[\vec{a}, \vec{b}, \vec{c}]+[\vec{a}, \vec{b}, \vec{a}]+[\vec{a}, \vec{c}, \vec{c}]+[\vec{a}, \vec{c}, \vec{a}]+[\vec{b}, \vec{b}, \vec{c}]+[\vec{b}, \vec{b}, \vec{a}]+[\vec{b}, \vec{c}, \vec{c}]+[\vec{b}, \vec{c}, \vec{a}] \)
\( =[\vec{a}, \vec{b}, \vec{c}]+0+0+0+0+0+0+[\vec{a}, \vec{b}, \vec{c}] \)
\( =2[\vec{a}, \vec{b}, \vec{c}] \)

\(\vec{b} \times \vec{c}=\left|\begin{array}{lrr} \hat{i} & \hat{j} & \hat{k} \\ -2 & 0 & 5 \\ 0 & +1 & -3 \end{array}\right|=-5 \hat{i}-6 \hat{j}-2 \hat{k}\)
\(\vec{a} \times(\vec{b} \times \vec{c})=\left|\begin{array}{llr} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ -5 & -6 & -2 \end{array}\right|=-12 \hat{i}+9 \hat{j}+3 \hat{k}\) ............(1)
\( \vec{a} \cdot \vec{c} =2(0)+3(1)+(-1)(-3)=6 \)
\(\vec{a} \cdot \vec{b} =2(-2)+3(0)+(-1)(5) \)
= -4 + 0 - 5 = -9
\( (\vec{a} \cdot \vec{c}) \vec{b} =-12 \hat{i}+30 \hat{k} \)
\((\vec{a} \cdot \vec{b}) \vec{c} =-9 \hat{j}+27 \hat{k}\)
\((\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=-12 \hat{i}+9 \hat{j}+3 \hat{k}\) .............(2)
From (1) & (2)
\(\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}\)

\( (\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d}) =(\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d})-(\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c}) \)
\(\vec{a} \cdot \vec{c} =2+1+1=4 \)
\(\vec{b} \cdot \vec{d} =2+0+2=4 \)
\(\vec{a} \cdot \vec{d} =1+1+2=4 \)
\(\vec{b} \cdot \vec{c} =4+1=5\)
L.H.S. = (4)(4) -(4)(5) = -4

\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 0 & 1 \end{array}\right|=\hat{i}+\hat{j}-2 \hat{k}\)
\(\vec{c} \times \vec{d}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{array}\right|=\hat{i}-3 \hat{j}+\hat{k}\)
\((\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \end{array}\right|=-5 \hat{i}-3 \hat{j}-4 \hat{k}\) .......(1)
\([\vec{a}, \vec{b}, \vec{c}]=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 0 & 1 \\ 2 & 1 & 1 \end{array}\right|=1\)
\([\vec{a}, \vec{b}, \vec{c}]=\left|\begin{array}{lll} 1 & 1 & 1 \\ 2 & 0 & 1 \\ 1 & 1 & 2 \end{array}\right|=-2\)
\( {[\vec{a}, \vec{b}, \vec{d}] \vec{c}-[\vec{a}, \vec{b}, \vec{c}] \vec{d} } =(-4 \hat{i}-2 \hat{j}-2 \hat{k})-(\hat{i}+\hat{j}+2 \hat{k}) \)
\( =-5 \hat{i}-3 \hat{j}-4 \hat{k}\) .........(2)
\( (1), (2) \Rightarrow(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=[\vec{a}, \vec{b}, \vec{d}] \vec{c}-[\vec{a}, \vec{b}, \vec{c}] \vec{d}\)

\((\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=\left|\begin{array}{ll} \vec{a} \cdot \vec{c} & \vec{a} \cdot \vec{d} \\ \vec{b} \cdot \vec{c} & \vec{b} \cdot \vec{d} \end{array}\right|=(\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d})-(\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{d})\)
\((\vec{b} \times \vec{c}) \cdot(\vec{a} \times \vec{d})=\left|\begin{array}{ll} \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{d} \\ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{d} \end{array}\right|=(\vec{b} \cdot \vec{a})(\vec{c} \cdot \vec{d})-(\vec{c} \cdot \vec{a})(\vec{b} \cdot \vec{d})\)
\((\vec{c} \times \vec{a}) \cdot(\vec{b} \times \vec{d})=\left|\begin{array}{ll} \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{d} \\ \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{d} \end{array}\right|=(\vec{c} \cdot \vec{b})(\vec{a} \cdot \vec{d})-(\vec{a} \cdot \vec{b})(\vec{c} \cdot \vec{d})\)
L.H.S
\( =(\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d})-(\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{d})+(\vec{b} \cdot \vec{a})(\vec{c} \cdot \vec{d})-(\vec{c} \cdot \vec{a})(\vec{b} \cdot \vec{d})+ (\vec{c} \cdot \vec{b})(\vec{a} \cdot \vec{d})-(\vec{a} \cdot \vec{b})(\vec{c} \cdot \vec{d})=0= \) R.H.S

Here p = 8 and \(\vec{n}=3 \hat{i}+2 \hat{j}-2 \hat{k}\)
\(\therefore \hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{3 \hat{i}+2 \hat{j}-2 \hat{k}}{\sqrt{9+4+4}}=\frac{3 \hat{i}+2 \hat{j}-2 \hat{k}}{\sqrt{17}}\)
Hence the required vector equation of the plane is
\( \vec{r} \cdot \hat{n} =p \)
\(\vec{r} \cdot \frac{3 \hat{i}+2 \hat{j}-2 \hat{k}}{\sqrt{17}} =8 \)
\(\vec{r} \cdot(3 \hat{i}+2 \hat{j}-2 \hat{k}) =8 \sqrt{17}\)
Cartesian form is \((x \hat{i}+x \hat{j}+z \hat{k}) \cdot(3 \hat{i}+2 \hat{j}-2 \hat{k})=8 \sqrt{17}\)
\(3 x+2 y-2 z=8 \sqrt{17}\)

The required plane Passes through the point  A(4, -2, -5) and is  Perpendicular to \(\overrightarrow{O A}\)
\(\therefore \vec{a}=4 \hat{i}-2 \hat{j}-5 \hat{k} \text { and } \vec{n}=\overrightarrow{O A}=4 \hat{i}-2 \hat{j}-5 \hat{k}\)
The required equation of the plane is \(\vec{r} \cdot \overrightarrow{\mathrm{n}}=\vec{a} \cdot \vec{n}\)
\(\vec{r} \cdot(4 \hat{i}-2 \hat{j}-5 \hat{k})=(4 \hat{i}-2 \hat{j}-5 \hat{k}) \cdot(4 \hat{i}-2 \hat{j}-5 \hat{k})\)
= 16 + 4 + 25
\(\vec{r} \cdot(4 \hat{i}-2 \hat{j}-5 \hat{k})=45\)
Cartesian form:
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(4 \hat{i}-2 \hat{j}-5 \hat{k})=45\)
4x -2y -5z = 45

The required plane passes through A(2, -1, -3) and parallel to \(\vec{u}=3 \hat{i}+2 \hat{j}-4 \hat{k} \text { and } \vec{v}=2 \hat{i}-3 \hat{j}+2 \hat{k}\)
The required equation is \(\vec{r}=\vec{a}+s \vec{u}+t \vec{v}\)
Cartesian form:
\( \left(x_{1}, y_{1}, z_{1}\right)=(2,-1,-3) ;\left(l_{1}, m_{1}, n_{1}\right)=(3,2,-4) ; \left(l_{2}, m_{2}, n_{2}\right)=(2,-3,2) \)
The equation of the plane is
 \(\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right|=0\)
\(\text { i.e., }\left|\begin{array}{ccc} x-2 & y+1 & z+3 \\ 3 & 2 & -4 \\ 2 & -3 & 2 \end{array}\right|=0\)
\(\Rightarrow 8 x+14 y+13 z+37=0\)
This is the required Cartesian form.

Vector equation of the plane passing through three given non-collinear points is
\(\vec{r}=(1-s-t) \vec{a}+s \vec{b}+t \vec{c}\) where s and t are scalars.
Here \( \vec{a}=2 \hat{i}+2 \hat{j}-\hat{k} ;\ \vec{b}=3 \hat{i}+4 \hat{j}+2 \hat{k} ;\ \vec{c}=7 \hat{i}+6 \hat{k} \)
\(\therefore \vec{r}=(1-s-t)(2 \hat{i}+2 \hat{j}-\hat{k})+s(3 \hat{i}+4 \hat{j}+2 \hat{k})+t(7 \hat{i}+6 \hat{k}) \)
Cartesian equation of the plane:
Here \( \left(x_{1}, y_{1}, z_{1}\right)\ is \ (2,2,-1) ;\left(x_{2}, y_{2}, z_{2}\right)\ is \ (3,4,2)\; \ \left(x_{3}, y_{3}, z_{3}\right)\ is \ (7,0,6)\)
The equation of the plane is
\(\left|\begin{array}{lll} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0\)
\(\text { i.e., }\left|\begin{array}{ccc} x-2 & y-2 & z+1 \\ 1 & 2 & 3 \\ 5 & -2 & 7 \end{array}\right|=0\)
5x + 2y - 32 = 17
This is the Cartesian equation of the plane.

Let \( \vec{a}=4 \hat{i}+2 \hat{j}+4 \hat{k} \)
\( \vec{b}=2 \hat{i}+5 \hat{j}+4 \hat{k} \)
\( \vec{d}=4 \hat{i}+7 \hat{j}+6 \hat{k} \)
\(\vec{b} \times \vec{d}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & 5 & 4 \\ 4 & 7 & 6 \end{array}\right|\)
\( =\hat{i}(30-28)-\hat{j}(12-16)+\hat{k}(14-20) \)
\( =2 \hat{i}+4 \hat{j}-6 \hat{k} \)
Non Parametric vector Equation
\( (\vec{r}-\vec{a}) \cdot(\vec{b} \times \vec{d})=0 \)
\((\vec{r}-(4 \hat{i}+2 \hat{j}+4 \hat{k})) \cdot(2 \hat{i}+4 \hat{j}-6 \hat{k})=0 \)
\( \vec{r} \cdot(2 \hat{i}+4 \hat{j}-6 \hat{k})-(8+8-24) =0 \)
\(\vec{r} \cdot(2 \hat{i}+4 \hat{j}-6 \hat{k})+8 =0 \)
\(\vec{r} \cdot(2 \hat{i}+4 \hat{j}-6 \hat{k}) =-8 \)
Cartesian Equation
\( (x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+4 \hat{j}-6 \hat{k}) =-8 \)
\(2 x+4 y-6 z =-8 \)
2x + 4y - 6z + 8 = 0

\( \frac{x-6}{1}=\frac{2-y}{2}=\frac{z-2}{2} \ and \ \frac{x-4}{3}=\frac{y}{-2}=\frac{1-z}{2} \)
\(\frac{x-6}{1}=\frac{y-2}{-2}=\frac{z-2}{2} \text { and } \frac{x+4}{3}=\frac{y}{-2}=\frac{z-1}{-2} \)
\(\vec{r}=(6 \hat{i}+2 \hat{j}+2 \hat{k})+t(\hat{i}-2 \hat{j}+2 \hat{k}) \)
\( \overrightarrow{\mathrm{r}}=(-4 \hat{\mathrm{i}}+\hat{\mathrm{k}})+\mathrm{s}(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \)
\( \vec{a}=6 \hat{i}+2 \hat{j}+2 \hat{k} ; \quad \vec{c}=-4 \hat{i}+\hat{k} \)
\( \vec{b}=\hat{i}+2 \hat{j}-2 \hat{k} ; \quad \vec{d}=3 \hat{i}+2 \hat{j}-2 \hat{k}\)
\(\delta=\frac{|(\vec{c}-\vec{a}) \cdot(\vec{b} \times \vec{d})|}{|(\vec{b} \times \vec{d})|}\)
\(\vec{b} \times \vec{d}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{array}\right|\)
\( =\hat{\mathrm{i}}(4+4)-\hat{\mathrm{j}}(-2-6)+\hat{\mathrm{k}}(-2+6) \)
\(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}} =8 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}(-2+6) \)
\(|\overrightarrow{\mathrm{b}} \times \mathrm{d}| =\sqrt{64+64+16}=\sqrt{144}=12\)
\( |(\vec{c}-\vec{a}) \cdot(\vec{b} \times d)| =|(-10 \hat{i}-6 \hat{j}-\hat{k}) \cdot(8 \hat{i}+8 \hat{j}+4 \hat{k})| \)
\(=|-80-48-4| \)
= 132

\( \vec{a}=\hat{i}+\hat{j}-\hat{k}, \vec{b}=2 \hat{i}+3 \hat{j}, \vec{c}=\hat{j}-\hat{k} \\ \vec{b} \times \vec{c} =\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ 0 & 1 & -1 \end{array}\right| \)
\( =\hat{i}(-3)-\hat{j}(-2)+\hat{\mathrm{k}}(2) \)
\(\vec{b} \times \vec{c} =-3 \hat{i}+2 \hat{j}+2 \hat{k} \)
\(\vec{a} \times(\vec{b} \times \vec{c})=\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ -3 & 2 & 2 \end{array}\right|\)
\( =\hat{i}(2+2)-\hat{j}(2-3)+\hat{k}(2+3) \)
\(=4 \hat{i}+\hat{j}+5 \hat{k}\)  .............(1)
\( \vec{a} \cdot \vec{c} =(\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{j}-\hat{k}) \)
\( =0+1+1=2 \)
\(=(\vec{a} \cdot \vec{c}) \vec{b}=4 \hat{i}+6 \hat{j} \)
\(\vec{a} \cdot \vec{b} =(\hat{i}+\hat{j}-\hat{k}) \cdot(2 \hat{i}+3 \hat{j}) \)
\( =2+3+0 \)
\(\vec{a} \cdot \vec{b} =5\)
\( (\vec{a} \cdot \vec{b}) \vec{c} =5 \hat{j}-5 \hat{k} \)
\((\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c} =(4 \hat{i}+6 \hat{j})-(5 \hat{j}-5 \hat{k}) \)
\( =4 \hat{j}+\hat{j}+5 \hat{k}\) ...............(2)
From (1) and (2), we get
\(\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}\)

The normal vector to the planes
x + 2y + 3z - 7 = 0, 2x - 3y + 4z = 0 are 
\(\overset { \rightarrow }{ b } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ c } =2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
∴ The required planes passes through the point \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \) and parallel to two vector 5 namely \(\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \) 
∴ The Parametric form of vectors equation of the plans is \(\overset { \rightarrow }{ r } =\overset { \rightarrow }{ a } +s\overset { \rightarrow }{ b } +t\overset { \rightarrow }{ c } \) s, t ∈ R
\(\overset { \rightarrow }{ r } =\left( \overset { \rightarrow }{ i } +\overset { \rightarrow }{ j } -\overset { \rightarrow }{ k } \right) +s\left( \overset { \rightarrow }{ i } +2\overset { \rightarrow }{ j } +3\overset { \rightarrow }{ k } \right) +t\left( 2\overset { \rightarrow }{ i } -3\overset { \rightarrow }{ j } +4\overset { \rightarrow }{ k } \right) ,\)
Cartesian equation is \(\left| \begin{matrix} x-{ x }_{ 1 } \\ { b }_{ 1 } \\ { c }_{ 1 } \end{matrix}\begin{matrix} y-{ { y }_{ 1 } } \\ { b }_{ 2 } \\ { c }_{ 2 } \end{matrix}\begin{matrix} z-{ { z }_{ 1 } } \\ { b }_{ 3 } \\ { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 \\ 1 \\ 2 \end{matrix}\begin{matrix} y-1 \\ 2 \\ -3 \end{matrix}\begin{matrix} z+1 \\ 3 \\ 4 \end{matrix} \right| =0\)
⇒ (x - 1) (8 + 9) - (y - 1)(4 - 6) + (z + 1)(-3 -4) = 0
⇒ 17 (x - 1) +2 (y - 1) -7 (z + 1) = 0
⇒ 17x - 17 + 2y - 2 - 7z - 7 = 0
⇒ 17x + 2y - 7z - 26 = 0

From the line  \(\frac { x-3 }{ 3 } =\frac { y-8 }{ -1 } =\frac { z-3 }{ 1 } \), we get
\(\overset { \rightarrow }{ a } =3\overset { \wedge }{ i } +8\overset { \wedge }{ j } +3\overset { \wedge }{ k } \), \(\overset { \rightarrow }{ b } =3\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
From the line \(\frac { x+3 }{ -3 } =\frac { y+7 }{ 2 } =\frac { z-6 }{ 4 } \) we get
\(\overset { \rightarrow }{ c } =-3\overset { \wedge }{ i } -7\overset { \wedge }{ j } +6\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ d } =-3\overset { \wedge }{ i } +2\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
Since the given lines are not parallel, the shortest distance between the line is 
\(d=\left| \frac { \left( \overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } \right) .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right) }{ \left| \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right| } \right| \)
\(\overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } =-6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } =\left| \begin{matrix} \overset { \wedge }{ i } \\ 3 \\ -3 \end{matrix}\begin{matrix} \overset { \wedge }{ j } \\ -1 \\ 2 \end{matrix}\begin{matrix} \overset { \wedge }{ k } \\ 1 \\ 4 \end{matrix} \right| \)
\(=\overset { \wedge }{ i } (-4-2)-\overset { \wedge }{ j } (12+3)+\overset { \wedge }{ k } (6-3)\\ \)
\(=-6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \)
\(\therefore \left| \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right| =\sqrt { 36+225+9 } \)
\(=\sqrt { 270 } \)
\(\therefore \left( \overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } \right) .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right) =\left( -6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) .\left( -6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) \)
= -6 (-6) + 15 _(15) + 3(3)
= 270 ≠ 0
Since the given lines are neither intersecting, nor parallel they are skew lines
\(\therefore d=\frac { 270 }{ \sqrt { 270 } } =\sqrt { 270 } units\)

Let \(\overset { \rightarrow }{ B } =x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \)
Given \(\overset { \rightarrow }{ A } \times \overset { \rightarrow }{ B } =\overset { \rightarrow }{ C } \Rightarrow \left| \begin{matrix} \overset { \wedge }{ i } \\ 1 \\ x \end{matrix}\begin{matrix} \overset { \wedge }{ j } \\ 1 \\ y \end{matrix}\begin{matrix} \overset { \wedge }{ k } \\ 1 \\ z \end{matrix} \right| =\overset { \wedge }{ j } -\overset { \wedge }{ k } \)
\(\Rightarrow \overset { \wedge }{ i } (z-y)-\overset { \wedge }{ j } (z-x)+\overset { \wedge }{ k } (y-x)\quad \overset { \wedge }{ j } -\overset { \wedge }{ k } \)
Equating the like components on both sides, we get
z - y = 0   .....(1)
x - y = 1   .....(2)
y - x = -1  .....(3)
Also, \(\overset { \rightarrow }{ A } .\overset { \rightarrow }{ B } =3\Rightarrow \left( \overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) .\left( x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \right) =3\)
⇒ x + y + z = 3   ....(4)
Solving (1), (2), (3) and (4), we get \(x=\frac { 5 }{ 3 } ,y=\frac { 2 }{ 3 } \)and \(z=\frac { 2 }{ 3 } \)
\(\therefore \overset { \rightarrow }{ B } =\frac { 5 }{ 3 } \overset { \wedge }{ i } +\frac { 2 }{ 3 } \overset { \wedge }{ j } +\frac { 2 }{ 3 } \overset { \wedge }{ k } \)

Given \(\overset { \rightarrow }{ AB } =\overset { \rightarrow }{ \alpha } \), \(\overset { \rightarrow }{ AD } =\overset { \rightarrow }{ \beta } \) and \(\overset { \rightarrow }{ AC } =2\overset { \rightarrow }{ \alpha } +3\overset { \rightarrow }{ \beta } \)
Area of the quadrilateral ABCD 
∴ = are of ∆ ABC + area of ∆ ACD
\(=\frac { 1 }{ 2 } \left| \overset { \rightarrow }{ AB } \times \overset { \rightarrow }{ AC } \right| +\frac { 1 }{ 2 } \left| \overset { \rightarrow }{ AC } \times \overset { \rightarrow }{ AD } \right| \)
\(=\frac { 1 }{ 2 } \left| \overset { \rightarrow }{ \alpha } \times \left( 2\overset { \rightarrow }{ \alpha } +3\overset { \rightarrow }{ \beta } \right) \right| +\frac { 1 }{ 2 } \left| \left( 2\overset { \rightarrow }{ \alpha } +3\overset { \rightarrow }{ \beta } \right) \times \overset { \rightarrow }{ \beta } \right| \)
\(=\frac { 1 }{ 2 } \left| 2\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \alpha } \right) +3\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) \right| +\frac { 1 }{ 2 } \left| 2\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) +3\left( \overset { \rightarrow }{ \beta } \times \overset { \rightarrow }{ \beta } \right) \right| \)
\(=\frac { 1 }{ 2 } \left| 3\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) \right| +\frac { 1 }{ 2 } \left| 2\left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) \right| \quad \quad \quad \left[ \because \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \alpha } =\overset { \rightarrow }{ \beta } \times \overset { \rightarrow }{ \beta } =0 \right] \)
\(=\left( \frac { 3 }{ 2 } +\frac { 2 }{ 2 } \right) \left( \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right) =\left( \frac { 5 }{ 2 } \right) \left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| \quad \quad (1)\)
Now, Area of the parallelogram with \(\overset { \rightarrow }{ AB } \) and \(\overset { \rightarrow }{ AD } \) as
adjacent sides = \(\left| \overset { \rightarrow }{ AB } \times \overset { \rightarrow }{ AD } \right| =\left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| .... (2)\)
From (1) & (2), \(\frac { 5 }{ 2 } \left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| =\lambda \left| \overset { \rightarrow }{ \alpha } \times \overset { \rightarrow }{ \beta } \right| \) [Given]
\(\lambda =\frac { 5 }{ 2 } \)

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.
\(\cos { A } =\frac { \overset { \rightarrow }{ AB } .\overset { \rightarrow }{ AC } }{ \left| \overset { \rightarrow }{ AB } \right| \left| \overset { \rightarrow }{ AC } \right| } \)
\(=\frac { \left( -\overset { \wedge }{ i } -2\overset { \wedge }{ j } -6\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ -i } -3\overset { \wedge }{ j } -5\overset { \wedge }{ k } \right) }{ \sqrt { 1+4+36 } .\sqrt { 1+9+25 } } \)
\(=\frac { 35 }{ \sqrt { 41 } .\sqrt { 35 } } =\sqrt { \frac { 35 }{ 41 } } =A={ Cos }^{ -1 }\left( \sqrt { \frac { 35 }{ 41 } } \right) \)
\(\cos { B } =\frac { \overset { \rightarrow }{ BA. } \overset { \rightarrow }{ BC } }{ \left| \overset { \rightarrow }{ BA } \right| \left| \overset { \rightarrow }{ BC } \right| } =\frac { \left( \overset { \wedge }{ i } +2\overset { \wedge }{ j } +6\overset { \wedge }{ k } \right) .\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) }{ \sqrt { 1+4+36 } +\sqrt { 4+1+1 } } \)
\(=\sqrt { \frac { 6 }{ 41 } } \Rightarrow B={ Cos }^{ -1 }\left( \sqrt { \frac { 6 }{ 41 } } \right) \)