Three Dimensional Geometry

To find the shortest distance between two skew lines given by:

$\mathbf{r} = (-\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 3\hat{\mathbf{k}}) + t(3\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 2\hat{\mathbf{k}})$

$\mathbf{r} = (7\hat{\mathbf{i}} + 4\hat{\mathbf{k}}) + s(\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + 2\hat{\mathbf{k}})$

follow these steps:

Define the Position Vectors and Direction Vectors:

$\mathbf{a}_1 = -\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 3\hat{\mathbf{k}}$

$\mathbf{b}_1 = 3\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 2\hat{\mathbf{k}}$

$\mathbf{a}_2 = 7\hat{\mathbf{i}} + 4\hat{\mathbf{k}}$

$\mathbf{b}_2 = \hat{\mathbf{i}} - 2\hat{\mathbf{j}} + 2\hat{\mathbf{k}}$

Calculate the Vector Connecting Points on the Lines:

$ \mathbf{a}_2 - \mathbf{a}_1 = (7 + 1)\hat{\mathbf{i}} + (0 + 2)\hat{\mathbf{j}} + (4 + 3)\hat{\mathbf{k}} = 8\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 7\hat{\mathbf{k}} $

Find the Cross Product of the Direction Vectors:

$ \mathbf{b}_1 \times \mathbf{b}_2 = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & -2 & -2 \\ 1 & -2 & 2 \end{vmatrix} $

$ = \hat{\mathbf{i}}((-2) \cdot 2 - (-2) \cdot (-2)) - \hat{\mathbf{j}}(3 \cdot 2 - (-2) \cdot 1) + \hat{\mathbf{k}}(3 \cdot (-2) - (-2) \cdot 1) $

$ = \hat{\mathbf{i}}(-4 - 4) - \hat{\mathbf{j}}(6 + 2) + \hat{\mathbf{k}}(-6 + 2) $

$ = -8\hat{\mathbf{i}} - 8\hat{\mathbf{j}} - 4\hat{\mathbf{k}} $

Calculate the Magnitude of the Cross Product:

$ \left|\mathbf{b}_1 \times \mathbf{b}_2\right| = \sqrt{(-8)^2 + (-8)^2 + (-4)^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12 $

Find the Required Distance:

$ \text{Distance} = \frac{|(8\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 7\hat{\mathbf{k}}) \cdot (-8\hat{\mathbf{i}} - 8\hat{\mathbf{j}} - 4\hat{\mathbf{k}})|}{12} $

$ = \frac{|8(-8) + 2(-8) + 7(-4)|}{12} $

$ = \frac{|-64 - 16 - 28|}{12} $

$ = \frac{108}{12} = 9 $

Thus, the shortest distance between these skew lines is 9.

Let $A E$ bisects $\angle B A C$.

$ \begin{aligned} A B & =\sqrt{(2-1)^2+(0-2)^2+(1-0)^2}=\sqrt{6} \\ A C & =\sqrt{(-3-1)^2+(0-2)^2+(2-0)^2} \\ & =\sqrt{24}=2 \sqrt{6} \\ \frac{B E}{E C} & =\frac{A B}{A C}=\frac{\sqrt{6}}{2 \sqrt{6}}=\frac{1}{2} \end{aligned} $

$\therefore E$ divides $B C$ in ratio 1:2

By section formula,

Coordinate of $E$ is

$ \begin{aligned} & \left(\frac{-3+4}{1+2}, \frac{0+0}{1+2}, \frac{2+2}{1+2}\right) \\ & \Rightarrow \\ & \begin{aligned} & A B\left.=\sqrt{\left(1-\frac{1}{3}, 0, \frac{4}{3}\right)}\right)^2+(2-0)^2+\left(0-\frac{4}{3}\right)^2 \\ &=\sqrt{\frac{4}{9}+4+\frac{16}{9}} \\ & \quad=\sqrt{\frac{56}{9}}=\frac{2 \sqrt{14}}{3} \end{aligned} \end{aligned} $

Equation of line joining $(0,2,4)$ and $(3,0,1)$ is

$ \frac{x}{3}=\frac{y-2}{-2}=\frac{z-4}{-3}=k $

Let coordinate of foot of perpendicular be $M$ ( $3 k, 2-2 k, 4-3 k$ )

Direction ratio of $B C=\langle 3,-2,-3\rangle$

Direction ratio of

$ A M=<3 k+1,1-2 k, 4-3 k\rangle $

As $A M$ and $B C$ are perpendicular

$ \begin{aligned} & 3(3 k+1)-2(1-2 k)-3(4-3 k)=0 \\ & 9 k+3-2+4 k-12+9 k=0 \\ & 22 k-11=0 \\ & \quad k=\frac{1}{2} \end{aligned} $

$\therefore$ Coordinate of $M$ is $\left(\frac{3}{2}, 1, \frac{5}{2}\right)$

$ \begin{aligned} & \text { Hence, required length }=A M \\ & =\sqrt{\left(-1-\frac{3}{2}\right)^2+(1-1)^2+\left(0-\frac{5}{2}\right)^2} \\ & =\sqrt{\frac{25}{4}+\frac{25}{4}}=\sqrt{2} \times \frac{5}{2}=\frac{5}{\sqrt{2}} \end{aligned} $

To find the relationship between the line and the plane, first determine their respective equations.

Equation of the Line:

The line $L$ passes through the points $(1, 2, -3)$ and $(\beta, 3, 1)$, but let's assume the correct coordinates at the second point are $(3, 3, -1)$ based on common calculation practices for vector direction. This gives the direction vector as $(3 - 1, 3 - 2, -1 + 3) = (2, 1, 2)$.

The parametric equations of the line can be written as:

$ \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z + 3}{2} $

Equation of the Plane:

The plane $\pi$ passes through the points $(2, 1, -2)$, $(-2, -3, 6)$, and $(0, 2, -1)$. The normal vector of the plane can be computed using the cross product of two vectors formed by these points:

Vector 1: $(-2 - 2, -3 - 1, 6 + 2) = (-4, -4, 8)$

Vector 2: $(0 - 2, 2 - 1, -1 + 2) = (-2, 1, 1)$

The cross product (normal vector) is:

$ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & -4 & 8 \\ -2 & 1 & 1 \end{vmatrix} = \mathbf{i}((1)(1) - (1)(8)) - \mathbf{j}((-4)(1) - (8)(-2)) + \mathbf{k}((-4)(1) - (-4)(-2)) $

Solving, we find:

$ \mathbf{i}(-4) - \mathbf{j}(12) + \mathbf{k}(-12) = (-4, -12, -12) $

Thus, the plane equation is:

$ -4(x - 2) - 12(y - 1) - 12(z + 2) = 0 $

Expanding and simplifying, we get:

$ -4x - 12y - 12z + 8 + 12 + 24 = 0 $

$ -4x - 12y - 12z + 44 = 0 $

It simplifies to:

$ x + y + z = 5 $

Angle Between Line and Plane:

The direction vector of the line is $(2, 1, 2)$ and the normal vector of the plane is $(1, 1, 1)$.

The angle $\theta$ between the line and the plane is given by:

$ \sin \theta = \frac{|2 \cdot 1 + 1 \cdot 1 + 2 \cdot 1|}{\sqrt{2^2 + 1^2 + 2^2} \sqrt{1^2 + 1^2 + 1^2}} = \frac{|2 + 1 + 2|}{\sqrt{9} \cdot \sqrt{3}} = \frac{5}{3\sqrt{3}} $

Now, compute $\cos^2 \theta$:

$ 27 \cos^2 \theta = 27 (1 - \sin^2 \theta) $

Substitute $\sin^2 \theta$:

$ 1 - \sin^2 \theta = 1 - \left(\frac{5}{3\sqrt{3}}\right)^2 = 1 - \frac{25}{27} = \frac{2}{27} $

Therefore,

$ 27 \cos^2 \theta = 27 \times \frac{2}{27} = 2 $

In conclusion, the value is $\boxed{2}$.

To determine if the points with the given position vectors are collinear, we need to set up vectors $ \overrightarrow{AB} $ and $ \overrightarrow{AC} $ as follows:

Let point $ A $ have the position vector $ \alpha \hat{\mathbf{i}} + 10 \hat{\mathbf{j}} + 13 \hat{\mathbf{k}} $.

Let point $ B $ have the position vector $ 6 \hat{\mathbf{i}} + 11 \hat{\mathbf{j}} + 11 \hat{\mathbf{k}} $.

Let point $ C $ have the position vector $ \frac{9}{2} \hat{\mathbf{i}} + \beta \hat{\mathbf{j}} - 8 \hat{\mathbf{k}} $.

The vector $ \overrightarrow{AB} $ is calculated as:

$ \overrightarrow{AB} = (6 - \alpha) \hat{\mathbf{i}} + (11 - 10) \hat{\mathbf{j}} + (11 - 13) \hat{\mathbf{k}} $

So:

$ \overrightarrow{AB} = (6 - \alpha) \hat{\mathbf{i}} + \hat{\mathbf{j}} - 2 \hat{\mathbf{k}} $

The vector $ \overrightarrow{AC} $ is calculated as:

$ \overrightarrow{AC} = \left(\frac{9}{2} - \alpha\right) \hat{\mathbf{i}} + (\beta - 10) \hat{\mathbf{j}} + (-8 - 13) \hat{\mathbf{k}} $

So:

$ \overrightarrow{AC} = \left(\frac{9}{2} - \alpha\right) \hat{\mathbf{i}} + (\beta - 10) \hat{\mathbf{j}} - 21 \hat{\mathbf{k}} $

Since the points are collinear, vectors $ \overrightarrow{AB} $ and $ \overrightarrow{AC} $ are parallel, which implies:

$ \frac{6 - \alpha}{\frac{9}{2} - \alpha} = \frac{1}{\beta - 10} = \frac{-2}{-21} $

Solving these equations, we find:

$ \frac{1}{\beta - 10} = \frac{2}{21} \; \Rightarrow \; \beta - 10 = \frac{21}{2} $

$ \beta = 10 + \frac{21}{2} = \frac{41}{2} $

For the other ratio:

$ \frac{12 - 2\alpha}{9 - 2\alpha} = \frac{2}{21} $

$ 126 - 21\alpha = 9 - 2\alpha $

So:

$ 19\alpha = 117 $

$ \alpha = \frac{117}{19} = 6 $

Given:

$ (19\alpha - 6\beta)^2 = (117 - 123)^2 = (-6)^2 = 36 $

Thus, the final result is $(19\alpha - 6\beta)^2 = 36$.

The given equation is $ axy + by = cy $.

We can simplify this to:

$ ax + bz = c $

This equation represents a plane. The key characteristic of this plane is its orientation in space.

Since there is no specific term dealing with the $ y $-coordinate in the rearranged equation $ ax + bz = c $, this plane is parallel to the $ Y $-axis.

Thus, it is a plane that is perpendicular to the $ ZX $-plane or could coincide with the $ ZX $-plane itself.

In simpler terms, the $ ZX $-plane is defined where $ y = 0 $, and any plane parallel to it or identical will have no effect from changes in the $ y $-coordinate.

We know that $Y \mathbb{Z}$ plane divides $P Q$ in the ratio 2:3.

By section formula the intersecting point is

$ \left(\frac{3 a+6}{5}, \frac{12+2 \beta}{5}, \frac{21+16}{5}\right) $

On $Y$ plane, $x$-coordinate of any point is zera.

$ \Rightarrow 3 \alpha+6=0 \Rightarrow \alpha=-2 $

$\therefore$ Coordinate of $P$ is $(-24,7)$

Also, $2 X$ plane divides $P Q$ in ratio $4: 5$.

The intersecting point is

$ \left(\frac{12+5 a}{9}, \frac{4 \beta+20}{9}, \frac{32+35}{9}\right) $

On $2 x$ plane, $y$-coordinate of any point is 0 .

$ \Rightarrow \quad 4 \beta+20=0 \Rightarrow \beta=-5 $

$\therefore$ Coordinate of $Q$ is $(3,-5.8)$

$ \begin{aligned} P Q & =\sqrt{13+2)^2+(-5-4)^2+(8-7)^2} \\ & =\sqrt{25+81+1} \\ & =\sqrt{107} \end{aligned} $

$A(1,2,3), B(3,-1,5), C(4,0,-3)$

Here,

$ \begin{aligned} & \mathbf{A B}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \mathbf{A C}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \end{aligned} $

$\mathbf{A B} \cdot \mathbf{A C}=6+6-12=0$

$\mathbf{A B} \perp \mathbf{A C} \Rightarrow$ triangle is right angled at $A$. Hence, circumcentre will be mid-point of $B$ and $C$.

Circumcentre $O\left(\frac{3+4}{2}, \frac{-1+0}{2}, \frac{5-3}{2}\right)$

$ \equiv O\left(\frac{7}{2}, \frac{-1}{2}, 1\right) $

$ \Rightarrow|\alpha|+|\beta|=\frac{7}{2}+\frac{1}{2}=4=4|\gamma| $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } l+m+n=0 \Rightarrow l=-(m+n) \\ & \Rightarrow \quad 2 l m+2 n l-m n=0 \\ & \Rightarrow \quad-2(m+n)^2-m n=0 \\ & \Rightarrow \quad 2 m^2+2 n^2+5 m n=0 \Rightarrow \frac{m}{n}=-2,-\frac{1}{2} \\ & \text { At } \frac{m}{n}=-2 \Rightarrow m=-2 n, l=n, n=n \\ & \Rightarrow 1,-2,1 \\ & \text { At } \frac{m}{n}=\frac{-1}{2} \Rightarrow n=-2 m, l=m, m=m \\ & \Rightarrow 1,1,-2 \\ & \cos \theta=\left|\frac{1-2-2}{\sqrt{1+4+1} \sqrt{1+1+4}}\right|=\frac{1}{2} \end{aligned} \end{aligned} $

We have, $a x+b y+c z=2$

Foot of perpendicular is given by

$ \frac{x-1}{a}=\frac{y-0}{b}=\frac{z+2}{c}=\frac{-(a+0-2 c-2)}{a^2+b^2+c^2} $

Given, foot of perpendicular is ( $2,0,-1$ )

$ \begin{aligned} & \Rightarrow \frac{2-1}{a}=\frac{0-0}{b}=\frac{-1+2}{c}=\frac{-(a-2 c-2)}{a^2+b^2+c^2} \\ & \Rightarrow \quad \frac{1}{a}=\frac{0}{b}=\frac{1}{c}=\frac{2 c-a+2}{a^2+b^2+c^2} \\ & \qquad b=0, a=c \\ & \text { and } \frac{1}{a}=\frac{2 a-a+2}{a^2+b^2+c^2} \\ & \Rightarrow \quad \frac{2 a^2}{a}=a+2 \Rightarrow a=2 \Rightarrow c=2 \\ & \therefore a^2+b^2+c^2=4+0+4=8 \end{aligned} $

Given as, $A(1,2,3), B(3,7,-2), C(6,7,7)$ and $D(-1,0,-1)$

Centroid of $\triangle A B D=(1,3,0) \rightarrow \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$

Centroid of $\triangle A C D=(2,3,3)$

$ \begin{aligned} \rightarrow & 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ \therefore \mathbf{r} & =\mathbf{a}+t(\mathbf{b}-\mathbf{a}) \\ \mathbf{r} & =(\hat{\mathbf{i}}+3 \hat{\mathbf{j}})+t(\hat{\mathbf{i}}+3 \hat{\mathbf{k}}) \\ \therefore \mathbf{r} & =(1+t) \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $

Let $A\left(x_1, y_1, z_1\right), B\left(x_2, y_2, z_2\right)$ and $C\left(x_3, y_3, z_3\right)$ are vertices of a triangle.

∵ Mid-point of $A B=(3,0,0)$

$ \therefore x_1+x_2=6, y_1+y_2=0, z_1+z_2=0 $

Similarly, $(0,4,0)$ and $(0,0,5)$ are mid-points of $B C$ and $C A$.

$ \begin{aligned} & \therefore \quad x_2+x_3=0, y_2+y_3=8, z_2+z_3=0 \\ & \text { and } x_1+x_3=0, y_1+y_3=0, z_1+z_3=10 \end{aligned} $

By solving these equations, we get

$ \begin{aligned} & x_1=3, y_1=-4, z_1=5, x_2=3, y_2=4 \\ & z_2=-5, x_3=-3, y_3=4, z_3=5 \end{aligned} $

So, $A(3,-4,5), B(3,4,-5)$ and $C(-3,4,5)$

$ \begin{aligned} & A B=\sqrt{0+(8)^2+(10)^2}=\sqrt{164} \\ & B C=\sqrt{(6)^2+0+(10)^2}=\sqrt{136} \end{aligned} $

and $C A=\sqrt{(-3-3)^2+(4+4)^2+0}=\sqrt{100}$

$ A B^2+B C^2+C A^2=164+136+100=400 $

Let $\mathbf{x}$ and $\mathbf{y}$ be two vectors whose direction cosines are $l, m, n$ and $a, b, c$ respectively.

$ \begin{aligned} & \therefore \quad \hat{\mathbf{x}}=l \hat{\mathbf{i}}+m \hat{\mathbf{j}}+n \hat{\mathbf{k}} \\ & \text { and } \hat{\mathbf{y}}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}=c \hat{\mathbf{k}} \end{aligned} $

Now, $\cos \theta=\hat{\mathbf{x}} \cdot \hat{\mathbf{y}}$

$ \Rightarrow \quad \hat{\mathbf{x}} \cdot \hat{\mathbf{y}}=l a+m b+n c $

For perpendicular $\hat{\mathbf{x}} \cdot \hat{\mathbf{y}}=0$

$ \therefore \quad l a+m b+n c=0 $

For parallel lines

$ \begin{array}{rlrl} & \hat{\mathbf{x}} =\lambda \hat{\mathbf{y}} \\ & \therefore l=\lambda a, \ m & =\lambda b \text { and } n=\lambda c \\ \Rightarrow & \frac{l}{a} =\frac{m}{b}=\frac{n}{c} \end{array} $

and the direction ratios of the lines bisecting the angle between is $l \pm a$, $m \pm b, n \pm c$.

Let $P(2,-1,3)$ be the foot of the perpendicular from the origin to the plane

$ \therefore \mathbf{O P}=\hat{\mathbf{n}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} $

and

$ \begin{aligned} & p=|2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}| \\ & =\sqrt{r^2+{1^2}^2+3^2}=\sqrt{4+1+9}=\sqrt{14} \end{aligned} $

$ \therefore \hat{\mathbf{n}}=\frac{2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{\sqrt{14}} $

The vector equation of the plane is

$ \begin{aligned} & \mathbf{r} \cdot \hat{\mathbf{n}}=p \\ & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot \frac{2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{\sqrt{14}}=\sqrt{14} \\ & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=14 \\ & 2 x-y+3 z=14 \\ & 2 x-y+3 z-14=0 \end{aligned} $

Given, $P$ divides $A B$ in the ratio $1: 2$.

∴ Coordinates of

$ P=\left(\frac{1 \times 2+2 \times 1}{1+2}, \frac{3 \times 1+2 \times 2}{1+2}, \frac{1 \times 1+3 \times 2}{1+2}\right) $

(by section formula)

$ \equiv\left(\frac{4}{3}, \frac{7}{3}, \frac{7}{3}\right) $

Coordinates of $Q$ divides $B C$ in the ratio -2:3.

∴ Coordinates of $Q$

$ \begin{aligned} & =\left(\frac{3 \times 2-2 \times 3}{1}, \frac{3 \times 3-2 \times 1}{1}, \frac{3 \times 1-4}{1}\right) \\ & =(0,7,-1) \end{aligned} $

Distance of $P Q$

$ \begin{aligned} & =\sqrt{\left(0-\frac{4}{3}\right)^2+\left(7-\frac{7}{3}\right)^2+\left(-1-\frac{7}{3}\right)^2} \\ & =\sqrt{\frac{16}{9}+\frac{196}{9}+\frac{100}{9}} \\ & =\sqrt{\frac{312}{9}}=\frac{2 \sqrt{78}}{3} \end{aligned} $

$ \begin{aligned} &\text { Equation of the line } B C\\ &\Rightarrow \quad \frac{x-2}{1}=\frac{y-2}{1}=\frac{z-1}{-1} \end{aligned} $

∵ Coordinates of point $M$

$ \equiv(\lambda+2, \lambda+2,-\lambda+1) $

Direction ratio of $A M$

$ \begin{array}{r} =(\lambda+2-1, \lambda+2+2,-\lambda+1-1) \\ =(\lambda+1, \lambda+4,-\lambda) \end{array} $

Now, $A M$ is perpendicular to $B C$

$ \begin{aligned} \therefore \lambda+1+\lambda+4+\lambda & =0 \\ 3 \lambda+5 & =0 \\ \Rightarrow \quad \lambda & =\frac{-5}{3} \end{aligned} $

∴ Coordinate of $M$

$ =\left(\frac{-5}{3}+2, \frac{-5}{3}+2, \frac{5}{3}+1\right)=\left(\frac{1}{3}, \frac{1}{3}, \frac{8}{3}\right) $

Now, coordinates of

$ \begin{aligned} & A^{\prime}=\left(\frac{2}{3}-1, \frac{2}{3}+2, \frac{16}{3}-1\right) \\ & \quad(l, m, n) \equiv\left(\frac{-1}{3}, \frac{8}{3}, \frac{13}{3}\right) \\ & l^2+m^2+n^2=\frac{1}{9}+\frac{64}{9}+\frac{169}{9}=\frac{234}{9}=26 \end{aligned} $

Let $A(1,1,1)$ is perpendicular to the line joining the points $P(6,3,2)$ and $Q(1,-4,-9)$

We know that direction ratios of a line ( $l$ )

$ \equiv\left(x_2-x_1, y_2-y_1, z_2-z_1\right) $

DR's of a line $\equiv(1-6,-4-3,-9-2)$

$ \equiv(-5,-7,-11) $

DR's of a normal (n) = DR's of a line

$ =(-5,-7,-11) $

Let $\left(a_1, b_1, c_1\right)=(-5,-7,-11)$

and $A(1,1,1) \equiv A\left(x_1, y_1, z_1\right)$

Equation of the plane is

$ \begin{aligned} & & a_1\left(x-x_1\right)+b_1\left(y-y_1\right)+c_1\left(z-z_1\right) & =0 \\ \Rightarrow & & -5(x-1)-7(y-1)-11(z-1) & =0 \\ \Rightarrow & & 5(x-1)+7(y-1)+11(z-1) & =0 \\ \Rightarrow & & 5 x+7 y+11 z-23 & =0 \end{aligned} $

On comparing the above equation with $a x+b y+c z-23=0$, we get

$ a=5, b=7 \text { and } c=11 $

Now, $a+b-c=5+7-11=1$

Equation of line passing through

$ \begin{aligned} & \hat{\mathbf{i}}-\hat{\mathbf{j}+0 \mathbf{k}, 0 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}} \\ & \frac{x-1}{0-1}=\frac{y+1}{1+1}=\frac{z-0}{-1-0} \\ & \frac{x-1}{-1}=\frac{y+1}{2}=\frac{z}{-1}=r\,\,\,\,\,\, (let)...(i)\end{aligned} $

Equation of plane passing through

$ \begin{aligned} & 2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+0 \hat{\mathbf{k}}, 0 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{i}}+0 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \qquad a(x-2)+b(y-1)+c(z-0)=0 \\ & \Rightarrow \quad-2 a+b-c=0 \\ & \text { and } \quad-a-b+2 c=0 \\ & \Rightarrow \quad 2 a-b+c=0 \\ & \text { and } \quad a+b-2 c=0 \end{aligned} $

$ \begin{aligned} a+b-2 c & =0 \\ \frac{a}{2-1} & =\frac{b}{1+4}=\frac{c}{2+1} \\ \frac{a}{1} & =\frac{b}{5}=\frac{c}{3} \end{aligned} $

$ \begin{aligned} & \therefore \quad(x-2)+5(y-1)+3(z-0)=0 \\ & \Rightarrow \quad x+5 y+3 z=7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

Point on line (i) is

$ \begin{aligned} &(1-r, 2 r-1,-r) \text { lies on plane (ii) } \\ & 1-r+5(2 r-1)+3(-r)=7 \\ & 1-r+10 r-5-3 r=7 \\ & \Rightarrow \quad r=11 / 6 \end{aligned} $

Point of intersection

$ \left(-\frac{5}{6}, \frac{16}{6},-\frac{11}{6}\right) $

or $\quad \frac{1}{6}(-5 \hat{\mathbf{i}}+16 \hat{\mathbf{j}}-11 \hat{\mathbf{k}})$

∵ Let the point $P(x, y, z)$ in the plane $\pi$.

Now, If plane $\pi$ is parallel to the plane having normal vector is $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$.

So, plane $\pi$ having normal vector as $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$

∴ Direction cosines of palne $\pi$ is $(x-3) \hat{\mathbf{i}}(y-4) \hat{\mathbf{j}}+(z-5) \hat{\mathbf{k}}$ with point $3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$.

So, $(x-3) \cdot 1+(y+4) 2+(z-5)(-3)=0$

$ \begin{array}{lr} \Rightarrow & x-3+2 y+8-3 z+15=0 \\ \therefore & x+2 y-3 z+20=0 \end{array} $

Given, lines

$ 3 l+2 m+n=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and  $ 2 m n-3 n l+5 l m=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eq. (i), we get

$ n=-(3 l+2 m) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

On putting in Eq. (ii), we get

$ \begin{array}{lr} & -2 m(3 l+2 m)+3 l(3 l+2 m)+5 l m=0 \\ \Rightarrow & -6 l m-4 m^2+9 l^2+6 l m+5 l m=0 \\ \Rightarrow & 9 l^2+5 l m-4 m^2=0 \\ \Rightarrow & 9 l^2+9 l m-4 l m-4 m^2=0 \\ \Rightarrow & 9 l(l+m)-4 m(l+m)=0 \\ \Rightarrow & (l+m)(9 l-4 m)=0 \\ \Rightarrow & l=-m \text { or } l=\frac{4 m}{9} \end{array} $

When $l=-m$, then from Eq. (iii), we get

$ \begin{aligned} & & n & =-(-3 m+2 m)=m \\ \Rightarrow & & -l & =m=n \\ \Rightarrow & & \frac{l}{-1} & =\frac{m}{1}=\frac{n}{1} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

When $l=\frac{4}{9} m$.

$ \begin{aligned} n & =-\left(\frac{4}{3} m+2 m\right)=-\frac{10}{3} m \\ \Rightarrow \quad \frac{l}{4 / 9} & =m=\frac{n}{-10 / 3} \\ \Rightarrow \quad \frac{l}{4} & =\frac{m}{9}=\frac{n}{-30} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\end{aligned} $

Angle between lines whose $\mathrm{DC}^{\prime}$ 's are proportional given in Eqs. (iv) and (v),

$ \cos \theta=\frac{(-1)(4)+(1)(9)+1(-30)}{\sqrt{1+1+1} \sqrt{16+81+900}}=\frac{-25}{\sqrt{2991}} $

Equation of plane parallel to

$ \begin{aligned} & x-y-z=0 \text { is } \\ & x-y-z+\lambda=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Plane Eq. (i) passes through ( $1,-2,1$ ).

$ \begin{aligned} \therefore & & 1+2-1+\lambda & =0 \\ \Rightarrow & & \lambda & =-2 \end{aligned} $

Equation of plane

$ x-y-z-2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On comparing with

$ a x+b y+c z-2=0 $

We have, $\quad a=1, b=-1, c=-1$

$ \begin{aligned} \therefore \quad b-2 c & =-1-2(-1) \\ & =-1+2=1=a \end{aligned} $

Now, first to find equation of plane passing through point $A(3,-2,1)$ and perpendicular to the vector $4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$.

$ \begin{aligned} & \text { So, } \mathbf{n} \cdot(4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})=(4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}) \\ & \cdot(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & \begin{aligned} & \Rightarrow(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}) \\ &=12-14-4=-6 \\ & 4 x+7 y-4 z=-6 \\ & \Rightarrow 4 x+7 y-4 z+6=0 \end{aligned} \end{aligned} $

Since, $M$ is the foot of the perpendicular drawn from $P(1,2,-1)$ to the plane $4 x+7 y-4 z+6=0$

So,

$ \begin{aligned} P M & =\left|\frac{4(1)+7(2)-4(-1)+6}{\sqrt{16+49+16}}\right| \\ & =\left|\frac{4+14+4+6}{\sqrt{81}}\right|=28 / 9 \end{aligned} $

Given, $A=(1,-1,2), B=(3,4,-2)$, $C=(0,3,2)$ and $D=(3,5,6)$

Now, DR's of the lines $\mathbf{A B}$ is $(2,5,-4)$ and DR's of the lines $\mathbf{C D}$ is $(3,2,4)$ Let ' $\theta$ ' be the angle between them, we get

$ \begin{aligned} \cos \theta & =\frac{\mathbf{A B} \cdot \mathbf{C D}}{|\mathbf{A B}| \cdot|\mathbf{C D}|} \\ & =\frac{6+10-16}{\sqrt{2^2+5^2+4^2} \sqrt{3^2+2^2+4^2}} \\ \Rightarrow \cos \theta & =0 \\ \therefore \quad \theta & =90^{\circ} \end{aligned} $

Assertion (A) DR'S of a line $L_1$ is 2, 5, 7 and

DR'S of a line $L_2$ is $\frac{4}{\sqrt{19}}, \frac{10}{\sqrt{19}}, \frac{14}{\sqrt{19}}$

Thus, $\frac{2}{\frac{4}{\sqrt{19}}}=\frac{5}{\frac{10}{\sqrt{19}}}=\frac{7}{\frac{14}{\sqrt{19}}}=\frac{1}{2}$

Hence, the lines $L_1$ and $L_2$ are parallel. Reason (R) If the DR'S of a line $L_1$ are $a_1, b_1, c_1$ and the $\mathrm{DR}^{\dagger} \mathrm{S}$ of a line $L_2$ are $a_2, b_2, c_2$ and $a_1 a_2+b_1 b_2+c_1 c_2=0$, then the lines are not parallel.

So, Assertion (A) is true but Reason (R) is false.

The given planes are $2 x+3 y+z=1$ and $x+3 y+2 z=2$

$ \begin{aligned} & \text { Thus, } \hat{\mathbf{n}}=\left|\begin{array}{lll} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{array}\right|=\hat{\mathbf{i}}(3)+\hat{\mathbf{j}}(-3)+\hat{\mathbf{k}}(3) \\ & =3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \text { and } \quad \hat{\mathbf{a}}=\frac{\mathbf{n}}{|\mathbf{n}|}=\frac{3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{\sqrt{9+9+9}}=\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}} \\ & \Rightarrow \quad \mathbf{a}=\frac{1}{\sqrt{3}} \hat{\mathbf{i}}-\frac{1}{\sqrt{3}} \hat{\mathbf{j}}+\frac{1}{\sqrt{3}} \hat{\mathbf{k}} \end{aligned} $

Since, the line ' $L$ ' makes an angle $\alpha$ with the positive direction of $X$-axis, then

$ \cos \alpha=\frac{1}{\sqrt{3}} $

Given the direction cosines $(l, m, n)$ of two lines, we have the following relationships:

$l + m + n = 0$

$lm = 0$

From the equation $lm = 0$, we infer that either $l = 0$ or $m = 0$.

Case 1: $l = 0$

Substitute $l = 0$ into the first equation: $m + n = 0$.

This gives $m = -n$.

Now, using the condition for direction cosines, $l^2 + m^2 + n^2 = 1$, we substitute $m = -n$ and $l = 0$:

$ 0^2 + (-n)^2 + n^2 = 1 \implies 2n^2 = 1 \implies n = \frac{1}{\sqrt{2}} $

Therefore, the set of direction cosines is:

$ \langle l, m, n \rangle = \left\langle 0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle $

Case 2: $m = 0$

Substitute $m = 0$ into the first equation: $l + n = 0$.

This gives $l = -n$.

Again using $l^2 + m^2 + n^2 = 1$ with $l = -n$ and $m = 0$:

$ (-n)^2 + 0^2 + n^2 = 1 \implies 2n^2 = 1 \implies n = \frac{1}{\sqrt{2}} $

Thus, the set of direction cosines is:

$ \langle l, m, n \rangle = \left\langle -\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}} \right\rangle $

Finding the Angle $\theta$ Between the Lines

The cosine of the angle $\theta$ between the two lines can be calculated using the dot product of their direction cosine vectors. Let us compute it:

$ \cos \theta = 0 \times \left(-\frac{1}{\sqrt{2}}\right) + \left(-\frac{1}{\sqrt{2}}\right) \times 0 + \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{2} $

Thus, the angle $\theta$ is:

$ \theta = \frac{\pi}{3} $

Perpendicular distance on $X$-axis

$ =y^2+z^2 $

Perpendicular distance on $Y$-axis

$ =x^2+z^2 $

Perpendicular distance on Z-axis

$ \begin{gathered} =x^2+y^2 \\ \therefore 2\left(x^2+y^2+z^2\right)=k\left(x^2+y^2+z^2\right) \end{gathered} $

Hence, $2=k$

Plane is,

$ \overline{\mathbf{r}} \cdot \overline{\mathbf{n}}=\overline{\mathbf{a}} \cdot \overline{\mathbf{n}} $

$ \begin{aligned} & \overline{\mathbf{r}} \cdot(-5 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+11 \hat{\mathbf{k}})=-\frac{15}{2}-\frac{21}{2}-\frac{99}{2} \\ \Rightarrow \quad & \quad-5 x-3 y+11 z=-\frac{135}{2} \\ \Rightarrow & -10 x-6 y+22 z+135=0 \\ \Rightarrow \quad & 10 x+6 y-22 z-135=0 \end{aligned} $

$\text { Any point on the line } \frac{x+1}{-3}=\frac{y-3}{2}=\frac{z-2}{-1} \text {, is of the form } (-3 \lambda-1,2 \lambda+3,-\lambda+2).$

Let the point $A \equiv(-3 \lambda-1,2 \lambda+3,-\lambda+2)$ and $B \equiv(-4,3,1)$

Then, $\mathbf{A B}=(3 \lambda-3) \hat{\mathbf{i}}-2 \lambda \hat{\mathbf{j}}+(\lambda-1) \hat{\mathbf{k}}$

Now, as the line is parallel to the given plane, therefore $\mathrm{AB}$ will be parallel to the given plane and so $\mathrm{AB}$ will be perpendicular to the normal of plane.

$\Rightarrow \mathrm{AB} \cdot \mathrm{n}=0$, where $\mathrm{n}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathbf{k}}$ is normal to the plane.

$\begin{aligned} & \Rightarrow(3 \lambda-3) \hat{\mathbf{i}}-2 \lambda \hat{\mathbf{j}}+(\lambda-1) \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})=0 \\ & \Rightarrow 3 \lambda-3-4 \lambda-\lambda+1=0 \\ & \Rightarrow-2 \lambda=2 \Rightarrow \lambda=-1 \end{aligned}$

Now, the required equation is the equation of line joining $A(2,1,3)$ and $B(-4,3,1)$, which is

$\begin{array}{ll} & \frac{x-(-4)}{2-(-4)}=\frac{y-3}{1-3}=\frac{z-1}{3-1} \\ \Rightarrow & \frac{x+4}{6}=\frac{y-3}{-2}=\frac{z-1}{2} \\ \text { or } & \frac{x+4}{3}=\frac{y-3}{-1}=\frac{z-1}{1} \end{array}$

$ \text { } \begin{aligned} & \mathbf{r}=(2-\lambda+\mu) \hat{\mathbf{i}}+(1-\mu) \hat{\mathbf{j}} +(2-3 \lambda+2 \mu) \hat{\mathbf{k}} \\ & \mathbf{r}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\lambda(-\hat{\mathbf{i}}-3 \hat{\mathbf{k}}) +\mu(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} $

which is a plane through $\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and parallel to the vectors.

$ \begin{aligned} & \mathbf{b}=-\hat{\mathbf{i}}-3 \hat{\mathbf{k}} \\ \text { and } & \mathbf{c}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned} $

Therefore, it is perpendicular to the vector.

$ \mathbf{n}=\mathbf{b} \times \mathbf{c} $

$ \begin{aligned} \mathbf{n} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -1 & 0 & -3 \\ 1 & -1 & 2 \end{array}\right| \\ & =\hat{\mathbf{i}}(0-3)-\hat{\mathbf{j}}(-2+3)+\hat{\mathbf{k}}(1-0) \\ \mathbf{n} & =-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} $

Hence, its vector equation is

$ \begin{aligned} & (\mathbf{r}-\mathbf{a}) \mathbf{n}=\mathbf{0} \\ \Rightarrow & \mathbf{r} \cdot \mathbf{n}=\mathbf{a} \cdot \mathbf{n} \\ \Rightarrow & \mathbf{r} \cdot(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=2(-3)+1(-1)+2(1) \\ & \mathbf{r} \cdot(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=-6-1+2 \\ & \mathbf{r} \cdot(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=-5 \end{aligned} $

Cartesian equation,

$ \begin{aligned} & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=-5 \\ \Rightarrow \quad & -3 x-y+z=-5 \\ \Rightarrow \quad & 3 x+y-z=5 \end{aligned} $

Option (a) is correct.

Equation of the passing through points

$(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$ and $(-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}})$ is

$ \mathbf{r}=(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+\lambda(4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}) $

Vector perpendicular to plane

$ \pi_1:(-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) $

Vector perpendicular to plane

$ \pi_2:(4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}) $

We know angle between two planes is same as angle between vectors perpendicular to them. So, let angle be $\theta$, then

$ \begin{aligned} \cos \theta & =\frac{(-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \cdot(4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}})}{|-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}||4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}|} \\ & =\frac{5}{\sqrt{5} \sqrt{41}}=\frac{\sqrt{5}}{\sqrt{41}}=\sqrt{\frac{5}{41}} \end{aligned} $

Given, $A=(1,2,0), B=(2,0,-1) C=(0,-2,3)$ and $D=(-1,2,-3)$. $G_1$ is centroid of $\triangle A B C$.

$ \begin{aligned} \therefore G_1 & =\left(\frac{1+2+0}{3}, \frac{2+0-2}{3}, \frac{0-1+3}{3}\right) \\ & =\left(1,0, \frac{2}{3}\right) \end{aligned} $

$G_2$ is centroid of tetrahedron $A B C D$.

$ G_2=\left(\frac{1+2+0-1}{4}, \frac{2+0-2+2}{4},\frac{0-1+3-3}{4}\right) $

$ =\left(\frac{1}{2}, \frac{1}{2}, \frac{-1}{4}\right) $

$ P(x, y, z)=\left(\frac{4 \times \frac{1}{2}+3 \times 1}{4+3}, \frac{4 \times \frac{1}{2}+3 \times 0}{4+3}\begin{aligned} , \left.\frac{4 \times\left(-\frac{1}{4}\right)+3\left(\frac{2}{3}\right)}{4+3}\right) \end{aligned}\right.$

$ \begin{aligned} & =\left(\frac{2+3}{7}, \frac{2+0}{7}, \frac{-1+2}{7}\right) \\ & =\left(\frac{5}{7}, \frac{2}{7}, \frac{1}{7}\right) \end{aligned} $

Given, $a-b+c=0$

$ \begin{array}{lr} \Rightarrow & b=a+c \,\,...(i)\\ \text { and } & a^2-b^2+2 c^2=0 \\ \Rightarrow & a^2-(a+c)^2+2 c^2=0 \end{array} $

[using Eq. (i)]

$ \begin{array}{lr} \Rightarrow & a^2-a^2-c^2-2 a c+2 c^2=0 \\ \Rightarrow & c^2-2 a c=0 \\ \Rightarrow & c(c-2 a)=0 \\ \Rightarrow & c=0 \text { or } c=2 a \end{array} $

When $c=0$, then $b=a$ and when $c=2 a$, then $b=3 a$

∴ Direction ratios of Eq. (ii) are $(1,1,0)$ and direction ratios of (iii) are ( $1,3,2$ ).

$ \begin{aligned} \therefore \cos \theta & =\frac{1 \times 1+1 \times 3+0 \times 2}{\sqrt{1^1+1^2+0} \sqrt{1^2+3^2+2^2}} \\ & =\frac{4}{\sqrt{2} \sqrt{14}}=\frac{2}{\sqrt{7}} \end{aligned} $

Let the given three point be $A \equiv(0,1,2), B \equiv(3,0,2)$ and $C \equiv(4,5,0)$ Equation of plane passing through three points is.

$ \begin{array}{rlrl} & \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-0 & y-1 & z-2 \\ 3-0 & 0-1 & 2-2 \\ 4-0 & 5-1 & 0-2 \end{array}\right| =0 \\ \Rightarrow & \left|\begin{array}{ccc} x & y-1 & z-2 \\ 3 & -1 & 0 \\ 4 & 4 & -2 \end{array}\right| =0 \end{array} $

$ \begin{array}{rlrl} & \Rightarrow & (z-2)[(3)(4)-(4)(-1)]-0 +(-2)[(x)(-1)-3(y-1)] & =0 \\ & \Rightarrow & (z-2)[12+4]-2[-x-3 y+3] & =0 \\ & \Rightarrow & 16 z-32+2 x+6 y-6 & =0 \\ & \Rightarrow & 2 x+6 y+16 z & =38 \\ & \Rightarrow & x+3 y+8 z & =19 \end{array} $

$ \begin{aligned} &\begin{aligned} \Rightarrow & \frac{x+3 y+8 z}{\sqrt{1+9+64}}=\frac{19}{\sqrt{1+9+64}} \\ & \frac{1}{\sqrt{74}} x+\frac{3}{\sqrt{74}} y+\frac{8}{\sqrt{74}} z=\frac{19}{\sqrt{74}} \end{aligned}\\ &\text { Comparing to } l x+m y+n z=p \text {, we have }\\ &\begin{array}{ll} & l=\frac{1}{\sqrt{74}}, m=\frac{3}{\sqrt{74}} \text { and } n=\frac{8}{\sqrt{74}} \\ \therefore & |l|+|m|+|n|=\frac{1+3+8}{\sqrt{74}}=\frac{12}{\sqrt{74}} \end{array} \end{aligned} $

Equation of line $L$ passing through points

$ \begin{aligned} & 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}} \text { and } \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \text { is } \\ & \mathbf{r}=(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}})+\lambda[(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}) \\ & \qquad \begin{array}{r} \quad-(\hat{\mathbf{i}}+6 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})] \\ L: \mathbf{r}=(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}})+\lambda(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \end{array} \\ & L: \frac{x-2}{1}=\frac{y-3}{-3}=\frac{z-8}{4}=\lambda \end{aligned} $

Now, plane is parallel to the vectors $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$

$ \mathbf{n}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 1 \\ 1 & -2 & 3 \end{array}\right|=-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}} $

So equation of plane passing through

$ -5 \hat{\mathbf{i}}+19 \hat{\mathbf{j}}-14 \hat{\mathbf{k}} $

$ \begin{aligned} a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right) & =0 \\ \Rightarrow-1(x+5)-2(y-19)-1(z+14) & =0 \\ P: x+2 y+z & =19 \end{aligned} $

Any general point on the line is

$ x=\lambda+2, y=-3 \lambda+3, z=4 \lambda+8 $

So, $A(\lambda+2,-3 \lambda+3,4 \lambda+8)$

Now, this point lies on the plane

$ \begin{gathered} \Rightarrow(\lambda+2)+2(-3 \lambda+3)+(4 \lambda+8)=19 \\ -\lambda=3 \Rightarrow \lambda=-3 \end{gathered} $

So, point of intersection

$ \begin{aligned} & A(\lambda+2,-3 \lambda+3,4 \lambda+8) \\ & A(-3+2,9+3,-12+8) \equiv(-1,12,-4) \end{aligned} $

So position vector of $A$ is $-\hat{\mathbf{i}}+12 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$.

Given, $P_1: \mathbf{r} \cdot(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})=5$

$ P_2: \mathbf{r} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=7 $

Let $\quad \mathbf{n}_1=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$

$ \mathbf{n}_2=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} $

Now, $\mathbf{n}=\mathbf{n}_1 \times \mathbf{n}_2$

$ \begin{aligned} \mathbf{n} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 3 & 4 \\ 1 & 1 & -1 \end{array}\right| \\ & =-7 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} $

Equation of line of intersection of planes passing through the point $(16,-9,0)$ is

$ \begin{aligned} \mathbf{r} & =(16 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+0 \hat{\mathbf{k}})+\lambda(-7 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & =(16-7 \lambda) \hat{\mathbf{i}}+(6 \lambda-9) \hat{\mathbf{j}}-\lambda \hat{\mathbf{k}} \end{aligned} $

Let $A(1,1,1), B(1,-4,3), C(2,-2,0)$ and $D(8,1,4)$ are the vertices of a tetrahedron.

Since, $G_1$ is the centroid of face $A B C$

$ \begin{aligned} \therefore G_1 & \equiv\binom{\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3},}{\frac{z_1+z_2+z_3}{3}} \\ G_1 & =\left(\frac{1+1+2}{3}, \frac{1-4-2}{3}, \frac{1+3+0}{3}\right) \\ G_1 & =\left(\frac{4}{3}, \frac{-5}{3}, \frac{4}{3}\right) \end{aligned} $

Now, $G_2$ is the centroid of face $B C D$,

$ \begin{aligned} \therefore G_2 & =\left(\frac{1+2+8}{3}, \frac{-4-2+1}{3}, \frac{3+0+4}{3}\right) \\ & G_2=\left(\frac{11}{3}, \frac{-5}{3}, \frac{7}{3}\right) \end{aligned} $

Also, $G_3$ is the centroid of the face $C D A$.

$ \begin{aligned} \therefore G_3 & =\left(\frac{1+2+8}{3}, \frac{1-2+1}{3}, \frac{1+0+4}{3}\right) \\ G_3 & \equiv\left(\frac{11}{3}, 0, \frac{5}{3}\right) \end{aligned} $

Also, $G_4$ is the centroid of the face $D A B$,

$ \begin{aligned} \therefore G_4 & =\left(\frac{1+1+8}{3}, \frac{1-4+1}{3}, \frac{1+3+4}{3}\right) \\ & G_4=\left(\frac{10}{3}, \frac{-2}{3}, \frac{8}{3}\right) \end{aligned} $

Now, centroid of tetrahedron having $G_1, G_2, G_3$ and $G_4$ as its vertices.

$ G=\left(\frac{\frac{4}{3}+\frac{11}{3}+\frac{11}{3}+\frac{10}{3}}{4}, \frac{-\frac{5}{3}-\frac{5}{3}+0-\frac{2}{3}}{4},\frac{\frac{4}{3}+\frac{7}{3}+\frac{5}{3}+\frac{8}{3}}{4}\right) $

$ \therefore G=(3,-1,2) $

$ \begin{aligned}Dr's\,\, of\,\,\, \mathbf{A B} & =\mathbf{B}-\mathbf{A} \\ & =(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}+0 \hat{\mathbf{k}})-(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & =2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned}Dr's \,\,of\,\, \mathbf{A} \mathbf{A C} & =\mathbf{C}-\mathbf{A} \\ & =(-\hat{\mathbf{i}}-\hat{\mathbf{j}}+11 \hat{\mathbf{k}})-(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & =-3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+12 \hat{\mathbf{k}} \end{aligned} $

If AD is the angle bisector of $\angle B A C$,

                   $ \begin{aligned} \mathbf{A D} & =\frac{\mathbf{A B}}{|\mathbf{A B}|}+\frac{\mathbf{A C}}{|\mathbf{A C}|} \\ & =\frac{2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}}{3}+\frac{-3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}}{13} \end{aligned} $

                          $ \begin{aligned} &=\frac{17 \hat{\mathbf{i}}-38 \hat{\mathbf{j}}+49 \hat{\mathbf{k}}}{3 \times 13}\\ &\text { Dr's : } \frac{17}{39}, \frac{-38}{39}, \frac{49}{39} \equiv(17,-38,49) \end{aligned} $

A plane passes through

$ A(2,3,0), B(0,-5,2) \text { and } C(-2,0,3) . $

Equation of plane passing through

$ \begin{aligned} & 3 \text { points }\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right)\left(x_3, y_3, z_3\right) \\ & \Rightarrow\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 \quad\left|\begin{array}{ccc} x-2 & y-3 & z-0 \\ 0-2 & -5-3 & 2-0 \\ -2-2 & 0-3 & 3-0 \end{array}\right|=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} x-2 & y-3 & z \\ -2 & -8 & 2 \\ -4 & -3 & 3 \end{array}\right|=0 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \\ & \Rightarrow \\ & \Rightarrow \end{aligned} \begin{aligned} &-18(x-2)-(y-3)(2)+z(-26)=0 \\ &\,\,\,\,\,\,9(x-2)+(y-3)+13 z=0 \\ & 9 x+y+13 z=21 \end{aligned}\\ &\text { If plane } P \text { cuts } X \text {-axis at } A \text {, then } A(x, 0,0)\\ &\begin{aligned} & \Rightarrow 9 x+0+0=21 \\ & \Rightarrow \quad x=7 / 3 \\ & \text { Point } A \equiv\left(\frac{7}{3}, 0,0\right) \end{aligned} \end{aligned} $

Given, the plane passing through the point

$ \begin{aligned} & \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, 3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}},-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \\ & (\mathbf{r}-\mathbf{a})[(\mathbf{b}-\mathbf{a}) \times(\mathbf{c}-\mathbf{a})]=0 \\ & \Rightarrow \mathbf{b}-\mathbf{a}=(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}})-(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \\ & =2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+7 \hat{\mathbf{k}} \\ & \Rightarrow \mathbf{c}-\mathbf{a}=(-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})-(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \\ & =-4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} & \therefore(\mathbf{b}-\mathbf{a}) \times(\mathbf{c}-\mathbf{a})=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & 7 \\ -4 & 4 & -2 \end{array}\right| \\ & =\hat{\mathbf{i}}(-2-28)-\hat{\mathbf{j}}(-4+28)+\hat{\mathbf{k}}(8+4) \\ & =-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}} \\ & \text { Now, }(\mathbf{r}-\mathbf{a}) \cdot[(\mathbf{b}-\mathbf{a}) \times(\mathbf{c}-\mathbf{a})]=0 \\ & \Rightarrow[\mathbf{r}-(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})][-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}]=0 \\ & \Rightarrow \mathbf{r}(-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}) +30-48+36=0 \end{aligned} $

$ \Rightarrow \mathbf{r}(-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}})=-18 $

From the options,

$ \begin{aligned}(a) & -\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} ;(-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ & (-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}) \\ & =30-72-24 \neq-18 \end{aligned} $

$ \begin{aligned}(b) & 7 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} ;(7 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}) \\ & (-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}) \\ & =-210+120-72 \neq-18 \end{aligned} $

$ \begin{aligned}(c) & -\hat{\mathbf{i}}+9 \hat{\mathbf{j}}+14 \hat{\mathbf{k}} ;(-\hat{\mathbf{i}}+9 \hat{\mathbf{j}}+14 \hat{\mathbf{k}}) \\ & (-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}) \\ & =30-216+168=-18 \end{aligned} $

$ \begin{aligned}(d) & 3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+8 \hat{\mathbf{k}} ;(3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}) \\ & (-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}) \\ & =-90+168+96 \neq-18 \end{aligned} $

Hence, the correct option is (c).

$ \begin{aligned} &\text { Given, angle between the planes }\\ &\begin{aligned} & \mathbf{r} \cdot(11 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}})=7 \text { and } \\ & \mathbf{r} \cdot(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})=5 \text { is } \pi / 2 \\ & \therefore \cos \frac{\pi}{2}=\frac{11 \times 2+(-2) \times 4-2 \alpha}{\sqrt{121+4+\alpha^2} \sqrt{4+16+4}} \end{aligned} \end{aligned} $

$ \begin{aligned} \Rightarrow & & 0 & =22-8-2 \alpha \\ \Rightarrow & & 2 \alpha & =14 \\ \Rightarrow & & \alpha & =7 \end{aligned} $

Given, $A(27,-243,81)$ is point in space.

$B, C$ and $D$ are the images of $A$ with respect to $X Y, Y Z$ and $Z X$ plane, respectively.

∴ Coordinate of $B, C$ and $D$ are $(27,-243,-81),(-27,-243,81)$ and (27, 243, 81), respectively.

Centroid of $B C D$ is

$ \begin{aligned} & \left(\frac{27-27+27}{3}, \frac{-243-243+243}{3},\right. \left.\frac{-81+81+81}{3}\right) \\ & =(9,-81,27) \\ & \therefore \alpha=9, \beta=-81 \text { and } \gamma=27 \\ & \alpha+\beta+\gamma=9-81+27=-45 \end{aligned} $

Given, a point $A(2,5,7)$ be the image of the point $B(1,-2,3)$, w.r.t. plane $\pi$.

Mid-point of $A B$ is $C$.

Coordinate of $C$ is $\left(\frac{3}{2}, \frac{3}{2}, 5\right)$.

DR's of $C D$ are $2-\frac{3}{2}, 1-\frac{3}{2}, 6-5$

$ =\frac{1}{2},-\frac{1}{2}, 1 $

DC's of $C D$ are

$ \begin{aligned} & =\frac{\frac{1}{2}}{\sqrt{\frac{1}{4}+\frac{1}{4}+1}}, \frac{-\frac{1}{2}}{\sqrt{\frac{1}{4}+\frac{1}{4}+1}}, \frac{1}{\sqrt{\frac{1}{4}+\frac{1}{4}+1}} \\ & =\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \end{aligned} $

Let the point $P$ divides the line $A B$ in the

ratio $\lambda: 1$.

Coordinate of $P$ is

$ \left(\frac{2 \lambda+1}{\lambda+1}, \frac{2 \lambda+1}{\lambda+1}, \frac{2 \lambda+1}{\lambda+1}\right) $

Point $P$ lies on the plane

$ \begin{aligned} & x+y+z-5=0 \\ & \Rightarrow \frac{2 \lambda+1}{\lambda+1}+\frac{2 \lambda+1}{\lambda+1}+\frac{2 \lambda+1}{\lambda+1}=5 \\ & \Rightarrow \quad 6 \lambda+3=5 \lambda+5 \\ & \Rightarrow \quad \lambda=2 \end{aligned} $

∴ The point $P$ divides line $A B$ in the ratio 2:1.

Hence, $A P: P B=2: 1$

Let position vector of

$ \mathbf{A}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}} $

Equation of line $L$ :

$ \frac{x-a}{2}=\frac{y-b}{1}=\frac{z-c}{-2} $

Point $P$ also lies on line $L$

$ \begin{aligned} & \frac{-7-a}{2}=\frac{-5-b}{1}=\frac{11-c}{-2}=\lambda(\text { let }) \\ & a=-7-2 \lambda, b=-5-\lambda \\and\,\, & c=11+2 \lambda \end{aligned} $

$ \begin{aligned} &\begin{aligned} \mathbf{A P} & =2 \lambda \hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}-2 \lambda \hat{\mathbf{k}} \\ |\mathbf{A P}| & =\sqrt{4 \lambda^2+\lambda^2+4 \lambda^2}=12 \\ 3 \lambda & =12 \Rightarrow \lambda=4 \\ \therefore \quad a & =-15, b=-9 \text { and } c=19 \end{aligned}\\ &\text { Positive vector of } \mathbf{A} \text { is }-15 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+19 \hat{\mathbf{k}} \end{aligned} $

Equation of bisector of no mals of the plane is

$ \begin{aligned} & \frac{4 x+3 y+\lambda}{\sqrt{4^2+3^2}}= \pm \frac{x+2 y+2 z+d}{\sqrt{1+4+4}} \\ & \frac{4 x+3 y+\lambda}{5}= \pm \frac{x+2 y+2 z+d}{3} \\ & 12 x+9 y+3 \lambda= \pm(5 x+10 y+10 z+5 d) \end{aligned} $

The equations are

$ \begin{aligned} & 7 x-y-10 z+3 \lambda-5 d=0 \\ & \text { and } 17 x+19 y+10 z+3 \lambda+5 d=0 \end{aligned} $

Hence, the bisector of normal is either along $17 \hat{\mathbf{i}}+19 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}$ and $7 \hat{\mathbf{i}}-\hat{\mathbf{j}}-10 \hat{\mathbf{k}}$.

Let coordinate of $D$ is $(a, b, c)$.

As $C$ is centroid of $A B C D$,

$ \begin{aligned} & G\left(\frac{1+2+3+a}{4},\right. \frac{2-3+2+b}{4} ,\left.\frac{3+1-1+c}{4}\right) \end{aligned} $

$ \begin{array}{ll} \Rightarrow & G\left(\frac{6+a}{4}, \frac{1+b}{4}, \frac{3+c}{4}\right) \\ \Rightarrow & \frac{6+a}{4}=\frac{5}{2}, \frac{1+b}{4}=\frac{3}{2}, \frac{3+c}{4}=\frac{9}{4} \\ \Rightarrow & a=4, b=5 \text { and } c=6 \end{array} $

∴ Coordinate of $G$ is $(4,5,6)$.

Let the point which divides $G D$ in the ratio $1: 2$ is ( $h, k, l$ ).

$ \begin{aligned} & \Rightarrow \quad h=\frac{1 \times 4+2 \times \frac{5}{2}}{1+2}, \\ & \quad k=\frac{1 \times 5+2 \times \frac{3}{2}}{1+2} \\ & \text { and } \quad l=\frac{1 \times 6+2 \times \frac{9}{4}}{1+2} \\ & \Rightarrow \quad h=3, k=\frac{8}{3} \text { and } l=\frac{7}{2} \end{aligned} $

∴ The required point is $\left(3, \frac{8}{3}, \frac{7}{2}\right)$.

$ \text { Let } D \text { divides } B C \text { in the ratio } k: 1 \text {. } $

Coordinate of $D$ is

$ \left(\frac{-2 k}{k+1}, \frac{4}{k+1}, \frac{4 k+1}{k+1}\right) $

Direction ratios of $B C=2,4,-3$

Direction ratios of $A D$ is $\frac{-2 k}{k+1}-2$,

$ \begin{aligned} & \frac{4}{k+1}, \frac{4 k+1}{k+1}-3 \\ & \quad=\frac{-4 k-2}{k+1}, \frac{4}{k+1}, \frac{k-2}{k+1} \end{aligned} $

As $A D \perp B C$,

$ \begin{gathered} 2\left(\frac{-4 k-2}{k+1}\right)+4\left(\frac{4}{k+1}\right)-3\left(\frac{k-2}{k+1}\right)=0 \\ -8 k-4 \pm 16-3 k+6=0 \\ -11 k+18=0 \Rightarrow k=18 / 11 \end{gathered} $

$\therefore D$ divides $B C$ in the ratio $18: 11$.

Given, equation of plane

$ \begin{aligned} & 6 x-3 y+2 z=6 \\ & \Rightarrow \quad x+\frac{y}{-2}+\frac{z}{3}=1 \\ & \therefore a=1, b=-2 \text { and } c=3 \end{aligned} $

Direction ratios of plane is $6,-3,2$.

Direction cosines of plane is

$ \begin{aligned} & \frac{6}{\sqrt{36+9+4}}, \frac{-3}{\sqrt{36+9+4}}, \frac{2}{\sqrt{36+9+4}} \\ & \therefore \quad l=\frac{6}{7}, m=\frac{-3}{7} \text { and } n=\frac{2}{7} \\ & \quad p=\left|\frac{0-0+0-6}{\sqrt{6^2+(-3)^2+2^2}}\right|=\frac{6}{7} \\ & \begin{aligned} |a l+b m+c n| & =\left|\frac{6}{7}+\frac{6}{7}+\frac{6}{7}\right|=3 \times \frac{6}{7} \\ & =3 p \end{aligned} \end{aligned} $