Let direction cosines of the line be I, m and n. 
Given \(\alpha=90^{\circ}, \beta=135^{\circ} \text { and } \gamma=45^{\circ}\) 
\(m=\cos \beta=\cos 135^{\circ}=\frac{-1}{\sqrt{2}}\) 
and \(n=\cos \gamma=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\) 
Hence, the direction cosines of a line are 0, \(\frac{-1}{\sqrt{2}} \text { and } \frac{1}{\sqrt{2}}\)

\(p=\frac { 70 }{ 11 } \)

Equation of the line is
\( 3\left(x+\frac{1}{3}\right)=6\left(y-\frac{1}{3}\right)=-1(z-1) \)
\(\Rightarrow \frac{x+\frac{1}{3}}{2}=\frac{y-\frac{1}{3}}{1}=\frac{z-1}{-6} \)
Line passes through fixed point \(\left(-\frac{1}{3}, \frac{1}{3}, 1\right)\) and direction ratios are 2, 1, - 6
Vector equation \(\vec { r } =\left( -\frac { 1 }{ 3 } \hat { i } +\frac { 1 }{ 3 } \hat { j } +\hat { k } \right) +\lambda (2\hat { i } +\hat { j } -6\hat { k } )\)

\(\frac { 25\sqrt { 14 } }{ 42 } \) units

General point on the line \( \frac{x-1}{3}=\frac{y}{2}=\frac{z+1}{7} \text { is }\)
\(Q(3 \lambda+1,2 \lambda, 7 \lambda-1)\)
If this point lies on the plane then
\(3 \lambda+1+2 \lambda-7 \lambda+1=8 \)
\(\Rightarrow-2 \lambda=6 \Rightarrow \lambda=-3 \)
Substituting in (i), point of intersection is
\(Q(-8,-6,-22)\)
Direction ratios of  P Q are12,12, 24 or 1, 1, 2
Equation of PQ is \(\frac { x-4 }{ 1 } =\frac { y-6 }{ 1 } =\frac { z-2 }{ 2 } \)

Let the d.c. 's of the lines be l, m, n.
Then \(\mathrm{I}=\cos 90^{\circ}=0, \mathrm{~m}=\cos 60^{\circ}=\frac{1}{2} n=\cos 30^{\circ}=\frac{\sqrt{3}}{2}\)

= 5 units

Let A(0, 3, 2), B(3, 5, 6)
Direction ratios of AB(a, b, c) are (3 - 0), (5 - 3), (6 - 2)
\((a_{ 1 },b_{ 1 },c_{ 1 })\) = (3, 2, 4)
Let C(1, - 1, 2), 0(3, 4, - 2)
Direction ratios of CD \((a_{ 2 },b_{ 2 },c_{ 2 })\) are (3 - 1), (4 + 1) (-2,2) \((a_{ 2 },b_{ 2 },c_{ 2 })\) are (2,56,-4)
When two lines are perpendicular if
\(a_{ 1 },a_{ 2 }+b_{ 1 },b_{ 2 },+c_{ 1 },c_{ 2 }=0\)
\(\Rightarrow 3\times 2+2\times 5+4\times -4=0\)
\(\Rightarrow 6+10-16=0\)
\(\Rightarrow 16-16=0\)
\(\therefore AB\bot to\quad CD\)

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.

Let points be A(2, 3, 4), B(- 1, - 2, 1) and C(5, 8, 7)
Direction ratios or AB are 2 + 1, 3 + 2, 4 - 1, i.e. 3, 5, 3;
Direction ratios of Be are 5 + 1, 8 + 2, 7 - 1, i.e. 3,5,3
As \(\frac{3}{3}=\frac{5}{5}=\frac{3}{3}\)
⇒  AB is parallel to BC, B is common.
Hence, A, B, C are collinear.

Given, equations of lines are
\(\begin{aligned} \vec{r}=(8 \hat{i}-19 \hat{j}+10 \hat{k})+\lambda(3 \hat{i}-16 \hat{j}+7 \hat{k}) \end{aligned}\)
and \(\begin{aligned} \vec{r}=(15 \hat{i}+29 \hat{j}+5 \hat{k})+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k}) \end{aligned}\)
On comparing with vector form of equation of a line,
i.e., \(\vec{r}=\vec{a}+\lambda \vec{b}\) we get
\(\vec{b}_1=3 \hat{i}-16 \hat{j}+7 \hat{k} \text { and } \vec{b}_2=3 \hat{i}+8 \hat{j}-5 \hat{k}\)
Now, we determine
\(\begin{aligned} \vec{b} & =\vec{b}_1 \times \vec{b}_2=\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{array}\right| \end{aligned}\)
\(=\hat{i}(80-56)-\hat{j}(-15-21)+\hat{k}(24+48) \)
\(=24 \hat{i}+36 \hat{j}+72 \hat{k}=12(2 \hat{i}+3 \hat{j}+6 \hat{k})\)
Since, the required line is perpendicular to the given lines. So, it is parallel to \(\vec{b}_1 \times \vec{b}_2\). Now, the equation of a line passing through the point (1, 2, -4) and parallel to \(24 \hat{i}+36 \hat{j}+72 \hat{k} \text { or }(2 \hat{i}+3 \hat{j}+6 \hat{k})\) is
\(\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})\)
which is required vector equation of a line.
For cartesian equation, put \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\), we get
\(x \hat{i}+y \hat{j}+z \hat{k}=(1+2 \lambda) \hat{i}+(2+3 \lambda) \hat{j}+(-4+6 \lambda) \hat{k}\)
On comparing the coefficients of \(\hat{i}, \hat{j} \text { and } \hat{k}\), we get
\(\begin{aligned} x=1+2 \lambda, y=2+3 \lambda \text { and } z=-4+6 \lambda \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{x-1}{2}=\lambda, \frac{y-2}{3}=\lambda \text { and } \frac{z+4}{6}=\lambda \end{aligned}\)
\(\begin{aligned} \therefore \quad \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6} \end{aligned}\)
which is the required cartesian equation of a line.
Alternate Method
Let the equation of line passing through (1, 2, -4) is
\(\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda_1\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)\)   ...(i)
Since, the line (i) is perpendicular to the given lines
\(\begin{aligned} \vec{r}=(8 \hat{i}-19 \hat{j}+10 \hat{k})+\lambda(3 \hat{i}-16 \hat{j}+7 \hat{k}) \end{aligned}\)
and  \(\begin{aligned} \vec{r}=(15 \hat{i}+29 \hat{j}+5 \hat{k})+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k}) \end{aligned}\)
Therefore, we have
\(\begin{aligned} \Rightarrow \quad\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right) \cdot(3 \hat{i}-16 \hat{j}+7 \hat{k})=0 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad 3 b_1-16 b_2+7 b_3=0 \end{aligned}\)                        ...(ii)
\(\begin{aligned} \text { and } \quad\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right) \cdot(3 \hat{i}+8 \hat{j}-5 \hat{k})=0 \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad 3 b_1+8 b_2-5 b_3=0 \end{aligned}\)                         ...(iii)
[\(\because\) if two lines \(\vec{r}=\vec{a}_1+\lambda \vec{b}_1 \text { and } \vec{r}=\vec{a}_2+\lambda \vec{b}_2\) are perpendicular, then \(\vec{b}_1 \cdot \vec{b}_2=0 .\)]
Now, on solving Eqs. (ii) and (iii), we get
\(\begin{array}{rlrl} \frac{b_1}{80-56}= \frac{b_2}{21+15}=\frac{b_3}{24+48} \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \frac{b_1}{24} & =\frac{b_2}{36}=\frac{b_3}{72} \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \frac{b_1}{2} & =\frac{b_2}{3}=\frac{b_3}{6} \end{array}\) [multiplying by 12]
\(\Rightarrow\) b₁ = 2k, b₂ = 3k and b₃ = 6k, for some constant k.
Thus, the required vector equation of line is
\(\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})\),
where \(\lambda=\lambda_1 k\) is any constant.
Now, for cartesian equation do same as in above method.

Given, L₁ : the line passing through (2, -1, 1) and parallel to \(\frac{x}{1}=\frac{y}{1}=\frac{z}{3}\)
\(L_2: \vec{r}=\hat{i}+(2 \mu+1) \hat{j}-(\mu+2) \hat{k}\)
Write the given equations of line in standard form
\(\begin{aligned} L_1: \vec{r}=(2 \hat{i}-\hat{j}+\hat{k})+\lambda(\hat{i}+\hat{j}+3 \hat{k})\end{aligned}\)     ....(i)
\(L_2: \vec{r}=(\hat{i}+\hat{j}-2 \hat{k})+\mu(2 \hat{j}-\hat{k})\)             ...(ii)
On comparing Eqs. (i) and (ii) with \(\vec{r}=a_1+\lambda b_1\) and \(\vec{r}=a_2+\lambda b_2\) respectively, we get
\(\vec{a}_1=2 \hat{i}-\hat{j}+\hat{k} \cdot \vec{b}_1=\hat{i}+\hat{j}+3 \hat{k}\)
and \(\vec{a}_2=(\hat{i}+\hat{j}-2 \hat{k}) \cdot \overrightarrow{b_x}=2 \hat{j}-\hat{k}\)
Clearly, \(\vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 3 \\ 0 & 2 & -1 \end{array}\right|\)
\(\begin{aligned} =\hat{i}(-1-6)-\hat{j}(-1-0)+\hat{k}(2-0) \end{aligned}\)
\(\begin{aligned} =-7 \hat{i}+\hat{j}+2 \hat{k} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad\left|\vec{b}_1 \times \vec{b}_2\right| & =|-7 \hat{i}+\hat{j}+2 \hat{k}| \end{aligned}\)
\(\begin{aligned} =\sqrt{(-7)^2+(1)^2+(2)^2} \end{aligned}\)
\(\begin{aligned} =\sqrt{49+1+4}=\sqrt{54}=3 \sqrt{6} \end{aligned}\)
Now, \(\begin{aligned} \overrightarrow{a_2}-\overrightarrow{a_1} & =(\hat{i}+\hat{j}-2 \hat{k})-(2 \hat{i}-\hat{j}+\hat{k}) \end{aligned}\)
\(\begin{aligned} =-\hat{i}+2 \hat{j}-3 \hat{k} \end{aligned}\)
Required SD \(=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\vec{b}_1 \times \vec{b}_2\right|}\)
\(\begin{aligned} =\frac{|(-\hat{i}+2 \hat{j}-3 \hat{k}) \cdot(-7 \hat{i}+\hat{j}+2 \hat{k})|}{3 \sqrt{6}} \end{aligned}\)
\(\begin{aligned} =\frac{|7+2-6|}{3 \sqrt{6}}=\frac{3}{3 \sqrt{6}}=\frac{1}{\sqrt{6}} \text { unit } \end{aligned}\)

The given equations can be rewritten as
= \(\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0} \text { and } \frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}\) 
= (4, 0, -1)

Converting the given equation of the line into cartesian form, we have
\(\frac { x+1 }{ 2 } =\frac { y-3 }{ 3 } =\frac { z-1 }{ -1 } =\lambda \)
Any point on the line (i) is :
\(Q(2\lambda -1,3\lambda +3,-\lambda +1)\)
For some values of A, let Q be perpendicular from P on the line (i).
Direction ratio's of PQ are:
\(2\lambda -6,3\lambda -1,-\lambda -1\)
Also, d.r.'s of line (i) are 2, 3, -1.
Since AM .Lt line (i), then
\(2(2\lambda -6)+3(3\lambda -1)-1(-\lambda -1)=0\)
\(\Rightarrow 14\lambda -14=0\)
\(\Rightarrow \lambda =1\)
\(\therefore \) From (ii), coordinates of foot of perpendicular Q are (10; 6, 0).
Now \(\bot \) distance PQ
\(=\sqrt { (1-5)^{ 2 }+(6-4)^{ 2 }+(0-2)^{ 2 } } \)
\(=\sqrt { 16+4+4 } =\sqrt { 24 } =2\sqrt { 6 } \)
Again, let the image of P in the line (i) is \(p^{ ' }(x_{ 1 },y_{ 1 },z_{ 1 })\) then is the mid-point of PP'.
\(\therefore \frac { 5+x_{ 1 } }{ 2 } ,=1\)
\(\frac { 4+y_{ 1 } }{ 2 } =6\)
\(\frac { 2+z_{ 1 } }{ 2 } =0\)
\(\therefore \quad x_{ 1 }=-3,y_{ 1 }=8,z_{ 1 }=-2\)
\(\therefore \) ImageofP in the line (i)is P' (-3, 5, -2).

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 equation of plane passing through three non - collinear points is given by
\(\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0\)
\(\Rightarrow\left|\begin{array}{ccc} x & y-1 & z-2 \\ 3-0 & 4-1 & -1-2 \\ 2-0 & 4-1 & 2-2 \end{array}\right|=0\)
\(\Rightarrow\left|\begin{array}{ccc} x & y-1 & z-2 \\ 3 & 3 & -3 \\ 2 & 3 & 0 \end{array}\right|=0\)
\( \Rightarrow \ln x(0+9)-(y-1)(0+6)+(z-2)(9-6)=0 \)
\(\Rightarrow 9 x-6 y+6+3 z-6=0 \Rightarrow 3 x-2 y+z=0\)
(ii) (c) : Height of tower = Perpendicular distance from the point (6, 5, 9) to the plane 3x - 2y + z = 0
\(=\left|\frac{18-10+9}{\sqrt{3^{2}+(-2)^{2}+1^{2}}}\right|=\frac{17}{\sqrt{14}} \text { units }\)
(iii) (b) : D.R.'s of perpendicular are < 3, -2, 1 >
[ஃ Perpendicular is parallel to the normal to the plane] Since, perpendicular is passing through the point (6, 5, 9), therefore its equation is 
\(\frac{x-6}{3}=\frac{y-5}{-2}=\frac{z-9}{1}\)
(iv) (a) : Let the coordinates of foot of perpendicular  are
Since, this point lie on the plane 3x - 2y + z = 0, therefore we get, \((3 \lambda+6,-2 \lambda+5, \lambda+9)\)
\( 3(3 \lambda+6)-2(-2 \lambda+5)+(\lambda+9)=0 \)
\(\Rightarrow 9 \lambda+4 \lambda+\lambda+18-10+9=0 \)
\(\Rightarrow 14 \lambda=-17 \Rightarrow \lambda=\frac{-17}{14}\)
Thus, the coordinates of foot of perpendicular are
\(\left(\frac{-51}{14}+6, \frac{34}{14}+5, \frac{-17}{14}+9\right) \text { i.e., }\left(\frac{33}{14}, \frac{104}{14}, \frac{109}{14}\right)\)
(v) (b) : Clearly, area of ABC =\(\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|\)
\( =\frac{1}{2}|(3 \hat{i}+3 \hat{j}-3 \hat{k}) \times(2 \hat{i}+3 \hat{j})| \)
\(=\frac{1}{2}\left\|\begin{array}{cc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & -3 \\ 2 & 3 & 0 \end{array}\right\|=\frac{1}{2}|9 \hat{i}-6 \hat{j}+3 \hat{k}|\)
\( =\frac{1}{2} \sqrt{9^{2}+6^{2}+3^{2}}=\frac{1}{2} \sqrt{126}=\frac{3}{2} \sqrt{14} \text { sq. units } \)

(i) (a) : Since, D.R.'s are proportional to D.C.'s, therefore L; will be perpendicular to L₂ iff
\(l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=0\)
(ii) (c) : Since, D.R.'s are proportional to D.C.'s, therefore L, will be parallel to L₂, iff
\(\frac{l_{1}}{l_{2}}=\frac{m_{1}}{m_{2}}=\frac{n_{1}}{n_{2}}\)
(iii) (c) : Equation of line joining Band C is \(\frac{x-1}{4}=\frac{y-4}{0}=\frac{z-6}{-2}\)
Let coordinates of foot of perpendicular be D(x, y, z).
ஃD.R.'s of AD are < x-1, y-2, z-1 >.
Now, 4(x - 1) + 0(y - 2) -2(z - 1) = 0 ⇒ 4x - 2z = 2
Also, (x, y, z) will satisfy equation of line BC.
Here, (3, 4, 5) satisfy both the conditions.
ஃ  Required coordinates are (3, 4, 5).
(iv) (b) : Let a, b, c be the direction ratios of the required line. Since it is perpendicular to the lines
whose direction ratios are (1, -2, -2) and (0, 2, 1) respectively.
ஃ  a - 2b - 2c = 0               ..... (i)
    0 .a + 2b + c = 0           .......(ii)
On solving (i) and (ii) by cross-multiplication, we get
\(\frac{a}{-2+4}=\frac{b}{0-1}=\frac{c}{2} \Rightarrow \frac{a}{2}=\frac{b}{-1}=\frac{c}{2}\)
Thus, the direction ratios of the required line are  < 2, -1, 2 >.
(v) (b) : D.R 's of given lines are < 3, -2, 0 > and < 1, -\(\frac{3}{2}\) ,2 >
Now, as 3.1 + (-2).(\(\frac{3}{2}\)) +0·2 = 3 -3 + 0 = 0
ஃ Given lines are perpendicular to each other.