Given line is \(\vec{r}=3 \hat{i}+5 \hat{j}-2 \hat{k}+\lambda(3 \hat{i}+\hat{j}+3 \hat{k})\) 
Equation of the plane passing through three points is
\(\left|\begin{array}{ccc} x-1 & y-1 & z \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{array}\right|=0\) 

\(\Rightarrow\) {(x-1)(-1-1)-(y-1)(0+3)+z(0+3)} = 0
\(\Rightarrow\) (x-1)(-2)-(y-1)(3)+3z = 0
\(\Rightarrow\) -2x+2-3y+3+3z = 0 or 2x+3y-3z-5 = 0
Since, \(2(3)+3(1)-3(3)=0 \Rightarrow\) lines is parallel to the plane.
Now, distance between the line and the plane
= distance of plane from any point on the line
\(=\left|\frac{2(3)+3(5)+(-3)(-2)-5}{\sqrt{(2)^{2}+(3)^{2}+(-3)^{2}}}\right|=\sqrt{22}\) \([\because(3,5,-2) \text { lies on the line }]\)

By triangle law of vector addition, we get
\(\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}\) 
 
\(\Rightarrow \overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=\overrightarrow{C A}+\overrightarrow{A C}\) 
\(\Rightarrow \overrightarrow{A B}+\overrightarrow{B C} \mid+\overrightarrow{C A}=\overrightarrow{C A}-\overrightarrow{C A} \quad[\because \overrightarrow{A C}=-\overrightarrow{C A}]\) 
\(\therefore \overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=\overrightarrow{0}\)

The angle between two planes is the angle between their normals. From the equation of the planes, the normal vectors are
\(\overrightarrow{\mathrm{N}}_{1}=2 \hat{i}+\hat{j}-2 \hat{k} \text { and } \overrightarrow{\mathrm{N}}_{2}=3 \hat{i}-6 \hat{j}-2 \hat{k}\)
\(\text {Therefore } \cos \theta=\left|\frac{\overrightarrow{\mathrm{N}}_{1} \cdot \overrightarrow{\mathrm{N}}_{2}}{\left|\overrightarrow{\mathrm{N}}_{1}\right|\left|\overrightarrow{\mathrm{N}}_{2}\right|}\right|=\left|\frac{(2 \hat{i}+\hat{j}-2 \hat{k}) \cdot(3 \hat{i}-6 \hat{j}-2 \hat{k})}{\sqrt{4+1+4} \sqrt{9+36+4}}\right|=\left(\frac{4}{21}\right)\)
\(\text {Hence }\theta=\cos ^{-1}\left(\frac{4}{21}\right) \)

Let line passing through \((\alpha, \beta, \gamma) \text { be }\)
\(\frac{x-\alpha}{a}=\frac{y-\beta}{b}=\frac{z-\gamma}{c}\)
According to given condition,
\(a l_{1}+b m_{1}+c n_{1}=0\)
\(\text { and } a l_{2}+b m_{2}+c n_{2}=0\)
\(\Rightarrow \frac{a}{m_{1} n_{2}-m_{2} n_{1}}=\frac{-b}{l_{1} n_{2}-l_{2} n_{1}}=\frac{c}{l_{1} m_{2}-l_{2} m_{1}}\)
Substitute in (i)
\(\text { Ans. } \frac{x-\alpha}{m_{1} n_{2}-m_{2} n_{1}}=\frac{y-\beta}{n_{1} l_{2}-n_{2} l_{1}}=\frac{z-\gamma}{l_{1} m_{2}-l_{2} m_{1}} \)

\(\overrightarrow{A B} =(4-3) \hat{i}+(x-2) \hat{j}+(5-1) \hat{k} \)
\(=\hat{i}+(x-2) \hat{j}+4 \hat{k} \)
\(\overrightarrow{A C} =(4-3) \hat{i}+(2-2) \hat{j}+(-2-1) \hat{k}=\hat{i}-3 \hat{k} \)
\(\overrightarrow{A D} =(6-3) \hat{i}+(5-2) \hat{j}+(-1-1) \hat{k} \)
\(=3 \hat{i}+3 \hat{j}+3 \hat{k} \)
If points are coplanar
\({[\overrightarrow{A B} \overrightarrow{A C} \overrightarrow{A D}]=0 \Rightarrow\left|\begin{array}{rrr} 1 & x-2 & 4 \\ 1 & 0 & -3 \\ 3 & 3 & -2 \end{array}\right|=0} \)
\(\Rightarrow 1(9)-(x-2)(7)+4(3)=0 \Rightarrow x=5 \)

\((\overrightarrow { a } -\overrightarrow { d } )\) x \((\overrightarrow { b } -\overrightarrow { c } )\) 
\(=\overrightarrow { a } *\overrightarrow { b } -\overrightarrow { a } *\overrightarrow { c } -\overrightarrow { d } *\overrightarrow { b } +\overrightarrow { d } *\overrightarrow { c } \)
\(=\overrightarrow { c } *\overrightarrow { d } -\overrightarrow { b } *\overrightarrow { d } -\overrightarrow { d } *\overrightarrow { b } +\overrightarrow { d } -\overrightarrow { c } \)
\([\because \overrightarrow { a } *\overrightarrow { b } =\overrightarrow { c } -\overrightarrow { d } and\overrightarrow { a } *\overrightarrow { c } =\overrightarrow { b } *\overrightarrow { d } ]\)
\(=\overrightarrow { c } *\overrightarrow { d } -\overrightarrow { b } *\overrightarrow { d } +\overrightarrow { b } *\overrightarrow { d } -\overrightarrow { c } *\overrightarrow { d } =\overrightarrow { 0 } \) 
Hence, \((\overrightarrow { a } -\overrightarrow { d } )\) is parallel to \((\overrightarrow { b } -\overrightarrow { c } )\)

Given, \(|\vec{a}|=3 \text { and } \vec{b}=\frac{2}{3}\)
Let \(\theta\) be the angle between \(\vec{a} \text { and } \vec{b}\).
Also, given \(|\vec{a} \times \vec{b}|=1\)
\(\begin{aligned} |\vec{a}||\vec{b}| \sin \theta=1 \end{aligned}\)
\(\Rightarrow \quad 3 \times \frac{2}{3} \sin \theta=1\)
\(\Rightarrow \quad 2 \sin \theta=1\)
\(\Rightarrow \quad \sin \theta=\frac{1}{2} \)
\(\Rightarrow \quad \theta=\frac{\pi}{6}\)

Distance of the point (3, - 5, 12) from the x-axis
\(=\sqrt{(-5)^2+(12)^2}=\sqrt{25+144} \)
\(=\sqrt{169}=13 \text { units }\)

Since, the required line is parallel to the line
\(\frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6} \)
or \( \frac{x+3}{3}=\frac{y-4}{-5}=\frac{z+8}{6} \)

Let Q divides PR in the ratio k: 1.Then, the coordinates of 
\(Q \operatorname{are}\left(\frac{9 k+3}{k+1}, \frac{8 k+2}{k+1}, \frac{-10 k-4}{k+1}\right)\) 
But it is given that coordinates of Q are (5,4, -6).
\(\therefore \quad \frac{9 k+3}{k+1}=5, \frac{8 k+2}{k+1}=4, \frac{-10 k-4}{k+1}=-6\) 
On solving all these equations, we get k = 1/2.
So, Q divides PR in the ratio 1: 2.

\(\overset\rightarrow a \times \overset\rightarrow b\)= \(\left| \begin{matrix} \overset { \wedge }{ i } & \overset { \wedge }{ j } & \overset { \wedge }{ k } \\ 1 & -1 & 2 \\ 2 & -1 & -1 \end{matrix} \right| \)
\(=i(1+2)-j(-1-4)+k(1-2)\)
= 3i + 5j + k
Area of \(\left| \overrightarrow { a } \times \overrightarrow { b } \right| =\sqrt{9+25+1}\)
\(=\sqrt{35} sq.units\)

\(\overset\rightarrow a=x\overset\wedge i+2\overset\wedge j-z\overset\wedge k \)
and \(\overset\rightarrow b=3\overset\wedge i-y\overset\wedge j+\overset\wedge k \)
are equal vectors
So, \(\overset\rightarrow a=\overset\rightarrow b\)
\(\Rightarrow x\overset\wedge i+2\overset\wedge j-z \overset\wedge k=3 \overset\wedge i-y\overset\wedge j+\overset\wedge k\)
\(\therefore \) x = 3 ,y = -2, z = -1
\(\therefore \) x + y + z = 3 - 2 - 1 = 0

(i) (b): The line along which motorcycle A is running, \(\vec{r}=\lambda(\hat{i}+2 \hat{j}-\hat{k})\) is which can be rewritten as \((x \hat{i}+y \hat{j}+z \hat{k})=\lambda \hat{i}+2 \lambda \hat{j}-\lambda \hat{k}\)
\(\Rightarrow x=\lambda, y=2 \lambda, z=-\lambda \Rightarrow \frac{x}{1}=\lambda, \frac{y}{2}=\lambda, \frac{z}{-1}=\lambda\)
Thus, the required cartesian equation is \(\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}\)
(ii) (d): Clearly, D.R:s of the required line are <  1, 2, -1  >
∴ D.Cs are \( <\frac{1}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}, \frac{2}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}, \frac{-1}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}> \)
\(\text { i.e., }<\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}>\)
(iii) (d): The line along which motorcycle B is running, is \(\vec{r}=(3 \hat{i}+3 \hat{j})+\mu(2 \hat{i}+\hat{j}+\hat{k})\), which is parallel to the vector \(2 \hat{i}+\hat{j}+\hat{k}\).
∴ D.R.'s of the required line are < 2, 1, 1 >.
(iv) (d): Here, \(\vec{a}_{1}=0 \hat{i}+0 \hat{j}+0 \hat{k}, \vec{a}_{2}=3 \hat{i}+3 \hat{j}, \vec{b}_{1}=\hat{i}+2 \hat{j}-\hat{k} \vec{b}_{2}=2 \hat{i}+\hat{j}+\hat{k}\)
\(\therefore \vec{a}_{2}-\vec{a}_{1}=3 \hat{i}+3 \hat{j}\)
and  \(\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 1 & 1 \end{array}\right|=3 \hat{i}-3 \hat{j}-3 \hat{k}\) 
Now, \(\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)=(3 \hat{i}+3 \hat{j}) \cdot(3 \hat{i}-3 \hat{j}-3 \hat{k})\)
= 9 - 9 = 0.
Hence, shortest distance between the given lines is 0.
(v) (c): Since, the point (1, 2, -1) satisfy both the equations of lines, therefore point of intersection of given lines is (1, 2, -1). So, the motorcycles will meet with an accident at the point (1, 2, -1).

Here, \(\left|\vec{F}_{1}\right|=\sqrt{(4)^{2}+0^{2}}=4 \mathrm{KN}\)
\( \left|\vec{F}_{2}\right|=\sqrt{(-2)^{2}+4^{2}}=\sqrt{20} \mathrm{KN} \)
\(\left|\vec{F}_{3}\right|=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{18} \mathrm{KN}\)
(i) (a): Since, \(\sqrt 20\) is larger. So, team B will win the game.
(ii) (b): Let F be the combined force
\(\therefore \vec{F}=\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}=4 \hat{i}+0 \hat{j}-3 \hat{i}-3 \hat{j}-2 \hat{i}+4 \hat{j} \)
\(=-\hat{i}+\hat{j} \)
\(\therefore |\vec{F}|=\sqrt{(-1)^{2}+1^{2}}=\sqrt{2}=1.4 \mathrm{KN}\)
(iii) (c) : We have,  \(\vec{F}=-\hat{i}+\hat{j}\)
\(\therefore \theta=\tan ^{-1}\left(\frac{F_{y}}{F_{x}}\right)=\tan ^{-1}\left(\frac{1}{-1}\right)=\frac{3 \pi}{4} \text { radian }\)
= 0.75 x 3.14 radian = 2.3555 radian ≈ 2.4 radian 
(iv) (a): Magnitude of force of Team B = \(\sqrt 20\) KN
= 2\(\sqrt 5\) KN
(v) (b): 4 KN force is applied by team A.

(a) Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.

(a) Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.

Given \(\vec{a} \perp(\vec{b}+\vec{c}), \vec{b} \perp(\vec{c}+\vec{a}), \vec{c} \perp(\vec{a}+\vec{b})\)
and \(|\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=5\)
To prove \(|\vec{a}+\vec{b}+\vec{c}|=5 \sqrt{2}\)
Consider, \(\begin{array}{r} |\vec{a}+\vec{b}+\vec{c}|^2=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c}) \end{array}\)
\(\begin{array}{r} {\left[\because|\vec{x}|^2=\vec{x} \cdot \vec{x}\right]} \end{array}\)
\(\begin{aligned} &=\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}+\vec{c} \cdot \vec{c} \end{aligned}\)
\(\begin{aligned} &=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+\vec{a} \cdot(\vec{b}+\vec{c})+\vec{b} \cdot(\vec{a}+\vec{c})+\vec{c} \cdot(\vec{a}+\vec{b}) \end{aligned}\)
\(=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+0+0+0\)
                                                \(\left[\begin{array}{l} \because \vec{a} \perp(\vec{b}+\vec{c}) \text {, therefore } \\ \vec{a} \cdot(\vec{b}+\vec{c})=0 \\ \text { Similarly, } \vec{b} \cdot(\vec{a}+\vec{c})=0 \\ \text { and } \vec{c} \cdot(\vec{a}+\vec{b})=0 \end{array}\right]\)
= 3² + 4² + 5² = 9 + 16 + 25          [given]
\(\Rightarrow|\vec{a}+\vec{b}+\vec{c}|^2=50 \Rightarrow|\vec{a}+\vec{b}+\vec{c}|=5 \sqrt{2}\)
                                                       [length cannot be '-' ve]

The vector equations of given lines are
\(\begin{aligned} \vec{r}=3 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(\hat{i}+2 \hat{j}+2 \hat{k}) \end{aligned}\)
and \(\begin{aligned} \vec{r}=5 \hat{i}-2 \hat{j}+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k}) \end{aligned}\)
On comparing them with \(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}\) and \(\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\), we get
\(\overrightarrow{a_1}=3 \hat{i}+2 \hat{j}-4 \hat{k}, \overrightarrow{a_2}=5 \hat{i}-2 \hat{j}, \vec{b}_1=\hat{i}+2 \hat{j}+2 \hat{k}\)
and \(\vec{b}_2=3 \hat{i}+2 \hat{j}+6 \hat{k}\)
\(\therefore \quad \overrightarrow{a_2}-\overrightarrow{a_1}=(5 \hat{i}-2 \hat{j})-(3 \hat{i}+2 \hat{j}-4 \hat{k})\)
\(=2 \hat{i}-4 \hat{j}+4 \hat{k}\)
\(\begin{aligned} \therefore \quad \vec{b}_1 \times \vec{b}_2 & =\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 3 & 2 & 6 \end{array}\right| \end{aligned}\)
\(\begin{aligned} =\hat{i}(12-4)-\hat{j}(6-6)+\hat{k}(2-6) \end{aligned}\)
\(\begin{aligned} =8 \hat{i}-4 \hat{k} \end{aligned}\)
\(\therefore\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)=(2 \hat{i}-4 \hat{j}+4 \hat{k}) \cdot(8 \hat{i}-4 \hat{k})\)
= 16 + 0 - 16 = 0
\(\therefore\) The lines are intersecting and the shortest distance between the lines is 0.
Now, the position vectors of arbitrary points on the given lines are \((3+\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(-4+2 \lambda) \hat{k}\) and \((5+3 \mu) \hat{i}+(-2+2 \mu) \hat{j}+6 \mu \hat{k}\), respectively.
Since, lines intersect then they have a common point.
\(\therefore \begin{aligned} 3+\lambda & =5+3 \mu \end{aligned}\)         ...(i)
\(\begin{aligned} 2+2 \lambda & =-2+2 \mu \end{aligned}\)        ...(ii)
\(\begin{aligned} -4+2 \lambda & =6 \mu \end{aligned}\)               ...(iii)
On solving Eqs. (i) and (ii), we get
\(\lambda=-4 \text { and } \mu=-2\)
\(\therefore\) Point of intersection is (3 - 4, 2 - 8, -4 - 8)
i.e. (-1, -6, -12).