DC's of required line are cos 60°, cos 120°and cos 45°;
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\text { i.e. } \frac{1}{2}, \frac{-1}{2}, \frac{1}{\sqrt{2}}\) and a point on the line is (2, -3, 4).
\(=\frac{x-2}{1 / 2}=\frac{y+3}{-1 / 2}=\frac{z-4}{1 / \sqrt{2}} \text { or } 2 x-4=-2 y-6=\sqrt{2}(z-4) \)

Given, cartesian equation of a line is \(\frac{3-x}{5}=\frac{y+4}{7}=\frac{2 z-6}{4}\)
On rewriting the given equation in standard form, we get
\(\frac{x-3}{-5}=\frac{y+4}{7}=\frac{z-3}{2}=\lambda\)  (let)
\(\Rightarrow \quad x=-5 \lambda+3, y=7 \lambda-4\)
and \(z=2 \lambda+3\)
Now, \(x \hat{i}+y \hat{j}+z \hat{k}=(-5 \lambda+3) \hat{i}+(7 \lambda-4) \hat{j}+(2 \lambda+3) \hat{k}\)
\(\therefore \quad \vec{r}=(3 \hat{i}-4 \hat{j}+3 \hat{k})+\lambda(-5 \hat{i}+7 \hat{j}+2 \hat{k})\)
which is the required equation of line in vector form.

The given lines are written as \(\frac{x-1}{-2}=\frac{y-4}{3 p}=\frac{z-3}{4} \text { and } \frac{x-2}{4 p}=\frac{y-5}{2}=\frac{z-1}{-7}\)
Here, direction ratios of above lines are (-2, 3p, 4) and (4p, 2, -7).
We know that two lines are perpendicular, if
a₁a₂ + b₁b₂ + c₁c₂ = 0
\(\begin{array}{ll} \therefore & -2 \times 4 p+3 p \times 2+4 \times(-7)=0 \end{array}\)
\(\begin{array}{ll} \Rightarrow & -8 p+6 p-28=0 \end{array}\)
\(\begin{array}{ll} \Rightarrow & -2 p=28 \Rightarrow p=-14 \end{array}\)
Hence, the value of p is -14.

We know that
\(cos^{ 2 }\alpha +cos^{ 2 }\beta +cos^{ 2 }\gamma =1\)
\(\Rightarrow (1-sin^{ 2 }\alpha )+(1-sin^{ 2 }\beta )+(1-sin^{ 2 }\gamma )\)
\(\Rightarrow sin^{ 2 }\alpha +sin^{ 2 }\beta +sin^{ 2 }\gamma =2\)

Point through which line passes is (2, 1, - 4) and dr's: 1, 1,-1.
\(\therefore\) Cartesian equation of line is \(\frac{z-2}{1}=\frac{y-1}{-1}=\frac{z+4}{-1}\)

AB

Direction ratios of line joining A and B are
\([1-2,-2-3,3-(-4)] \text { , i.e. }(-1,-5,7)\) 
Direction ratios ofline joining Band Care
3 –1, 8 + 2, – 11 – 3, i.e., 2, 10, – 14
It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel to BC. But point B is common to both AB and BC. Therefore, A, B, C are collinear points.

Let the vertices of \(\Delta A B C\)  be A(3, 5, -4), B(-1, 1, 2) and C( -5, -5, -2).
Then, the direction ratios of side AB are
[(-1-3), (1-5), 2-(- 4)1 Le.(-4, -4, 6).
[\(\therefore\) if the given points are A(x₁, y₁, z₁) and B(x₂, y₂, z₂),then DR's of AB = (x₂ - x₁, y₂ - y_1, z₂ - z₁)] and the direction cosines of AB are
\(\left(\frac{-4}{2 \sqrt{17}}, \frac{-4}{2 \sqrt{17}}, \frac{6}{2 \sqrt{17}}\right) \text { or }\left(\frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}\right)\) 
if a, b and c are direction ratios, then dirction cosines are 
\(\left(\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}\right)\) 
Similarly, the direction ratios of side BC are \([\{-5-(-1)\},(-5-1),(-2-2)], \text { i.e. }(-4,-6,-4)\) 
and the direction cosines of BC are
\(\left(\frac{-4}{2 \sqrt{17}}, \frac{-6}{2 \sqrt{17}}, \frac{-4}{2 \sqrt{17}}\right) \text { or }\left(\frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}\right)\) 
The direction ratios of side AC are
\([(-5-3),(-5-5),\{(-2-(-4)\}], \text { i.e. }(-8,-10,2)\) 
and the direction cosines of AC are
\(\left(\frac{-8}{2 \sqrt{42}}, \frac{-10}{2 \sqrt{42}}, \frac{2}{2 \sqrt{42}}\right) \text { or }\left(\frac{-4}{\sqrt{42}}, \frac{-5}{\sqrt{42}}, \frac{1}{\sqrt{42}}\right)\)

Any line through(1, 2 ,-4) is:
\(\vec { r } .\left( \hat { i } +2\hat { j } -4\hat { k } \right) +\lambda ({ b }_{ 1 }\hat { i } +{ b }_{ 2 }\hat { j } +{ b }_{ 3 }\hat { k } )..(i)\)
\(where\ { b }_{ 1 }\hat { i } +{ b }_{ 2 }\hat { j } +{ b }_{ 3 }\hat { k } \) is its direction.
\(The\ line\ \frac { x-8 }{ 3 } =\frac { y+19 }{ -16 } =\frac { z+10 }{ 7 } ...(ii)\)
has direction \(3 \hat { i } -16\hat { j } +7\hat { k } \)
Now lines (1) and (2) are perpendicular,
\(\therefore \left( { b }_{ 1 }\hat { i } +{ b }_{ 2 }\hat { j } +{ b }_{ 3 }\hat { k } \right) .\left( \hat { i } -16\hat { j } +7\hat { k } \right) =0\)
\(\Rightarrow 3{ b }_{ 1 }-16{ b }_{ 2 }+7{ b }_{ 3 }=0....(iii)\)
\(\text {Similarly }3{ b }_{ 1 }+8{ b }_{ 2 }-5{ b }_{ 3 }=0.....(iv)\)
\(\text {Do it Solving (3) and (4),}\)
\(\frac { { b }_{ 1 } }{ 80-56 } =\frac { { b }_{ 2 } }{ 21+15 } =\frac { { b }_{ 3 } }{ 24+48 } \)
\(\Rightarrow \frac { { b }_{ 1 } }{ 24 } =\frac { { b }_{ 2 } }{ 36 } =\frac { { b }_{ 3 } }{ 72 } \Rightarrow \frac { { b }_{ 1 } }{ 2 } =\frac { { b }_{ 2 } }{ 3 } =\frac { { b }_{ 3 } }{ 6 } \)
The direction of (i) is \( 2\hat { i } +3\hat { j } +6\hat { k } \)
The equation of line (i) is \( \vec { r } .\left( \hat { i } +2\hat { j } -4\hat { k } \right) +\lambda (2\hat { i } +3\hat { j } +6\hat { k } )\)

\(\left| \frac { -16-36-64 }{ \sqrt { 16+36+64 } } \right| =\left| \frac { -116 }{ \sqrt { 116 } } \right| =2\sqrt { 29 } \) units

Plane passing through the point A(0, 0, 0) is
a(x - 0) + bev - 0) + c(z - 0) = 0 ...(i)
Plane (i) passes through the point (3, -1, 2)
3a - b + 2c = 0
Plane (i) is parallel to the line \( \frac{\dot{x}-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}\)
\(\therefore \quad a-4 b+7 c=0\)
Eliminating a, b, c  from (i), (ii) and (iii), we get
\(\left|\begin{array}{lll} x & y & z \\ 3 & -1 & 2 \\ 1 & -4 & 7 \end{array}\right|=0 \)
\(\Rightarrow x(-7+8)-y(21-2)+z(-12+1)=0 \)
\(\Rightarrow x-19 y-11 z=0 \) is the required equation.

Any point on the given line is \(\left( 2\lambda -1,\ -2\lambda +3,\ -\lambda \right) \). In this point is Q then
\(\vec { PQ } =\left( 2\lambda -2 \right) \hat { i } +\left( -2\lambda +1 \right) \hat { j } +\left( -\lambda +3 \right) \hat { k } \)
Since \(\vec { PQ }\) is perpendicular to the line 
\(\therefore 2\left( 2\lambda -2 \right) -2\left( -2\lambda +1 \right) -1\left( -\lambda +3 \right) =0\)
\(\Rightarrow \lambda =1\)
 \(\therefore\)  Foot of perpendicular is Q(1, 1, -1)
Let P'(x, y, z) be the image of P in the line, then
\(\frac { x+1 }{ 2 } =1\),
\(\frac { y+2 }{ 2 } =1\),
\(\frac { y-3 }{ 2 } \) =-1
\(\Rightarrow\) x = 1, y = 0, z = 1
\(\Rightarrow\) Image is (1, 0, 1)

The lines which are neither intersecting nor parallel are called skew lines.
The given vector equations of two skew lines are
\(\vec{r}=8 \hat{i}-9 \hat{j}+10 \hat{k}+\lambda(3 \hat{i}-16 \hat{j}+7 \hat{k})\)       ...(i)
and \(\vec{r}=15 \hat{i}+29 \hat{j}+5 \hat{k}+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k})\) ...(ii)
Here, \(\vec{a}_1=8 \hat{i}-9 \hat{j}+10 \hat{k} ; \vec{a}_2=15 \hat{i}+29 \hat{j}+5 \hat{k}\)
and \(\overrightarrow{b_1}=3 \hat{i}-16 \hat{j}+7 \hat{k} ; \quad \overrightarrow{b_2}=3 \hat{i}+8 \hat{j}-5 \hat{k}\)
Now, \(\overrightarrow{a_2}-\overrightarrow{a_1}=(15-8) \hat{i}+(29+9) \hat{j}+(5-10) \hat{k}\)
\(\begin{aligned} &=7 \hat{i}+38 \hat{j}-5 \hat{k} \end{aligned}\)
\(\begin{aligned} \therefore \quad \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{array}\right| \end{aligned}\)
\(\begin{aligned} =\hat{i}(80-56)-\hat{j}(-15-21)+\hat{k}(24+48) \end{aligned}\)
\(\begin{aligned} =24 \hat{i}+36 \hat{j}+72 \hat{k} \end{aligned}\)
\(\begin{aligned} \text { and }\left(\vec{b}_1\right. & \left.\times \vec{b}_2\right) \cdot\left(\overrightarrow{a_2}-\vec{a}_1\right) \end{aligned}\)
\(\begin{aligned} =(24 \hat{i}+36 \hat{j}+72 \hat{k}) \cdot(7 \hat{i}+38 \hat{j}-5 \hat{k}) \end{aligned}\)
= 168 + 1368 - 360 = 1176
\(\therefore\) Required shortest distance
\(\begin{aligned} d & =\left|\frac{\left(\overrightarrow{b_1} \times \vec{b}_2\right) \cdot\left(\overrightarrow{a_2}-\vec{a}_1\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right| \end{aligned}\)
\(\begin{aligned} =\left|\frac{1176}{\sqrt{24^2+36^2+72^2}}\right| \end{aligned}\)
\(\begin{aligned} =\frac{1176}{\sqrt{7056}} \end{aligned}\)
\(\begin{aligned} d & =\frac{1176}{84}=14 \text { units } \end{aligned}\)

As 3 x \(\frac{k^2}{9}\) = 1 ⇒ k 土\(\sqrt3\)

As on the y-axis, x-coordinate and z-coordinate are zeroes

As \({ cos }^{ 2 }\alpha +{ cos }^{ 2 }\beta +{ cos }^{ 2 }\gamma =1\)
\(\Rightarrow \frac { 1+cos2\alpha }{ 2 } +\frac { 1+cos2\beta }{ 2 } +\frac { 1+cos2\gamma }{ 2 } =1\)
= cos 2α + cos 2β + cos 2γ = 1

(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).