3D Geometry

Equation of plane passing through point (1, 2, 2) is,

a(x $-$ 1) + b(y $-$ 2) + c(z $-$ 2) = 0        . . .(1)

This plane is perpendicular to the plane

x $-$ y + 2z = 3   and  2x $-$ 2y + z + 12 = 0

When two planes,

a₁x + b₁y + c₁z + d₁ = 0

and a₂x + b₂y + c₂z + d₂ = 0

are perpendicular to each other then

a₁a₂ + b₁b₂ + c₁c₂ = 0

So, we can say,

a $ \times $ 1 + b $ \times $ ($-$ 1) + c $ \times $ 2 = 0

$ \Rightarrow $   a $-$ b + 2c = 0           . . . (2)

and, a $ \times $ 2 + b($-$2) + c(1) = 0

$ \Rightarrow $   2a $-$ 2b + c = 0           . . .(3)

Solving (2) and (3)

${a \over { - 1 + 4}}$ = ${b \over {4 - 1}}$ = ${c \over { - 2 + 2}}$ = $\lambda $

$ \Rightarrow $    a = 3$\lambda $, b = 3$\lambda $, c = 0

Putting the values of a, b, c in equation (1) we get,

3$\lambda $ (x $-$ 1) + 3$\lambda $ (y $-$ 2) = 0

$ \Rightarrow $   3(x $-$ 1) + 3(y $-$ 2) = 0

$ \Rightarrow $   3x + 3y $-$ 9 = 0

$ \Rightarrow $   x + y $-$ 3 = 0

$ \therefore $   Distance of point (1, $-$2, 4) from plane x + y $-$ 3 = 0 is,

= $\left| {{{1 - 2 - 3} \over {\sqrt {1 + 1} }}} \right|$ = ${4 \over {\sqrt 2 }}$ = 2$\sqrt 2 $

$e{q^n}\,\,$ of $\,\,PO:\,\,$ ${{x - 1} \over 1} = {{y + 5} \over 1} = {{z - 9} \over 1} = \lambda $

$ \Rightarrow x = \lambda + 1;\,\,y = \lambda - 5;\,\,z = \lambda + 9$

Putting these in $e{q^n}$ of plane :-

$\lambda + 1 - \lambda + 5 + \lambda + 9 = 5$

$ \Rightarrow \lambda = - 10$

$ \Rightarrow O$ is $\left( { - 9, - 15, - 1} \right)$

$ \Rightarrow $ $\,\,\,$ Distance $OP = 10\sqrt 3 .$

Line lies in the plane $ \Rightarrow \left( {3, - 2, - 4} \right)$ lie in the plane

$ \Rightarrow 3\ell - 2m + 4 = 9$ or $3\ell - 2m = 5....\left( 1 \right)$

Also, $\ell ,$ $m, - 1$ are dr's of line perpendicular to plane

and $2, - 1,3$ are dr's of line lying in the plane

$ \Rightarrow 2\ell - m - 3 = 0\,\,\,$ or $\,\,\,2\ell - m = 3....\left( 2 \right)$

Solving $(1)$ and $(2)$ we get $\ell = 1$ and $m=-1$

$ \Rightarrow {\ell ^2} + {m^2} = 2.$

Let P(x₁, y₁, z₁) image of Q(3, 1, 7) w.r.t. the plane x $-$ y + z = 3.

Let R be the point on plane which is midpoint of the line joining P and Q.

The equation of the line PQ is

${{x - 3} \over 1} = {{y - 1} \over { - 1}} = {{z - 7} \over 1} = \lambda $

Therefore, (x, y, z) = (3 + $\lambda$, 1 $-$ $\lambda$, 7 + $\lambda$) lies on plane.

$3 + \lambda - 1 + \lambda + 7 + \lambda = 3$

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

The point R is (3 $-$ 2, 3, 7 $-$ 2) = (1, 3, 5).

Now, ${{{x_1} + 3} \over 2} = 1 \Rightarrow {x_1} = - 1$

${{{y_1} + 1} \over 2} = 3 \Rightarrow {y_1} = 5$

${{{z_1} + 7} \over 2} = 5 \Rightarrow {z_1} = + 3$

That is, the point P is P($-$1, 5, +3).

Now, the equation of the plane passing through P is

$a(x + 1) + b(y - 5) + c(z - 3) = 0$

This plane contains the line:

${x \over 1} = {y \over 2} = {z \over 1}$

This is, $a(0 + 1) + b(0 - 5) + c(0 - 3) = 0 \Rightarrow a = 5b + 3c$

$a + 2b + c = 0 \Rightarrow 7b + 4c = 0 \Rightarrow b = {{ - 4c} \over 7}$

$a = - {{20c} \over 7} + 3c = {c \over 7}$

Now, the equation of the plane is obtained as follows:

${c \over 7}(x + 1) - {{4c} \over 7}(y - 5) + c(z - 3) = 0$

$(x + 1) - 4(y - 5) + 7(z - 3) = 0$

$x + 1 - 4y + 20 + 7z - 21 = 0$

$x - 4y + 7z = 0$

Let us consider a pyramid OPQRS in the first octant with O as origin as shown in the following figure:


Now, the midpoint of $O Q$ is $T\left(\frac{3}{2}, \frac{3}{2}, 0\right)$.

Also, it is obvious that the point $S$ is given by $S\left(\frac{3}{2}, \frac{3}{2}, 3\right)$.

Now, $O T=\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^2+0}=\frac{3}{2} \sqrt{2}$

Therefore, $\cos \theta=\frac{3}{2} \frac{\sqrt{2}}{3}=\frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4}$

where $\theta$ is the angle between $O Q$ and $O S$, which is calculated as follows:

The ratio of the plane containing triangle $\triangle O Q S$ can be written as

$ \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 0 \\ 3 / 2 & 3 / 2 & 3 \end{array}\right|=i(9)-i(9)+k(0)=i(i-j) $

Therefore, the ratio is $(1,-1,0)$.

Now, the equation of the plane is

$ \begin{aligned} &1(x-0)-1(y-0)+0(z-0) =0 \\\\ &x-y =0 \end{aligned} $

Hence, option (B) is correct.

The coordinates of the length of perpendicular from point $P$ to the plane $x-y=0$ is $(3,0,0)$.

Hence, the length of perpendicular from point $P$ to the plane containing the triangle $O Q S$ is

$ \frac{z-0}{\sqrt{2}}=\frac{3}{\sqrt{2}} $

Hence, option (C) is correct.

Now, points $R$ and $S$, respectively, are $R(0,3,0)$ and $ S\left(\frac{3}{2}, \frac{3}{2}, 3\right) \text {. } $

The equation of line $R S$ is

$ \begin{aligned} \bar{r} & =3 \hat{j}+\lambda\left(\frac{3}{2} \hat{i}+\left(\frac{3}{2}-3\right) \hat{j}+3 \hat{k}\right) \\\\ & =3 \hat{j}+\lambda\left(\frac{3}{2} \hat{i}-\frac{3}{2} \hat{j}+3 \hat{k}\right) \end{aligned} $


Let point A be the feet of perpendicular from O to line RS:

$ \begin{aligned} & A\left(3 \hat{j}+\mu\left(\frac{3}{2} \hat{i}-\frac{3}{2} \hat{j}+3 \hat{k}\right)\right) \\\\ & \text {Now, } \overrightarrow{O A}=3 \hat{j}+\mu\left(\frac{3}{2} \hat{i}-\frac{3}{2} \hat{j}+3 \hat{k}\right) \end{aligned} $

That is, $(\vec{b}-\vec{a})=\left(\frac{3}{2} \hat{i}-\frac{3}{2} \hat{j}+3 \hat{k}\right)$

Therefore,

$ \overrightarrow{O A} \cdot(\vec{b}-\vec{a})=0 $

$ \begin{aligned} &\left(3 \hat{j}+\mu\left(\frac{3}{2} \hat{i}-\frac{3}{2} \hat{j}+3 \hat{k}\right)\right) \cdot\left(\frac{3}{2} \hat{i}-\frac{3}{2} \hat{j}+3 \hat{k}\right)=0 \\\\ &\frac{-9}{2}+\mu\left(\frac{9}{4}+\frac{9}{4}+9\right)=0 \\\\ &\frac{-9}{2}+\mu\left(\frac{9}{2}+9\right)=0 \\\\ &\Rightarrow-\frac{1}{2}+\mu\left(\frac{3}{2}\right)=0 \\\\ &\Rightarrow \mu=\frac{1}{3} \end{aligned} $

Therefore, $A\left(3 \hat{j}+\frac{\hat{i}}{2}-\frac{\hat{j}}{2}+\hat{k}\right)=A\left(\frac{\hat{i}}{2}+\frac{5 \hat{j}}{2}+\hat{k}\right)$ and hence the perpendicular distance from point $O$ to the straight line containing the line $R S$ is

$ |\overrightarrow{O A}|=\sqrt{\frac{1}{4}+\frac{25}{4}+\frac{4}{4}}=\sqrt{\frac{30}{4}}=\sqrt{\frac{15}{2}} $

Hence, option (D) is correct.

General point on given line $ \equiv P\left( {3r + 2,4r - 1,12r + 2} \right)$

Point $P$ must satisfy equation of plane

$\left( {3r + 2} \right) - \left( {4r - 1} \right) + \left( {12r + 2} \right) = 16$

$11r + 5 = 16$

$r=1$

$P\left( {3 \times 1 + 2,4 \times 1 - 1,12 \times 1 + 2} \right) = P\left( {5,3,14} \right)$

Distance between $P$ and $(1,0,2)$

$D = \sqrt {{{\left( {5 - 1} \right)}^2} + {3^2} + {{\left( {14 - 2} \right)}^2}} = 13$

Equation of the plane containing the lines

$2x - 5y + z = 3$ and $x + y + 4z = 5$ is

$2x - 5y + z - 3 + \lambda \left( {x + y + 4z - 5} \right) = 0$

$ \Rightarrow \left( {2 + \lambda } \right)x + \left( { - 5 + \lambda } \right)y + \left( {1 + 4\lambda } \right)z + $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { - 3 - 5\lambda } \right) = 0....\left( i \right)$

Since the plane $(i)$ parallel to the given plane

$x + 3y + 6z = 1$

$\therefore$ $\,\,\,{{2 + \lambda } \over 1} = {{ - 5 + \lambda } \over 3} = {{1 + 4\lambda } \over 6}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \lambda = - {{11} \over 2}$

Hence equation of the required plane is

$\left( {2 - {{11} \over 2}} \right)x + \left( { - 5 - {{11} \over 2}} \right)y + \left( {1 - {{44} \over 2}} \right)z + $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { - 3 + {{55} \over 2}} \right) = 0$

$ \Rightarrow x + 3y + 6z = 7$

Let the equation of the line passing through the origin be $\mathrm{L}: \frac{x}{l}=\frac{y}{m}=\frac{z}{n}\quad \text{... (i)}$

Given, planes

$\begin{aligned} & \mathrm{P}_1: x+2 y-z+1=0 \quad \text{... (ii)}\\ & \mathrm{P}_2: 2 x-y+z-1=0 \quad \text{... (iii)} \end{aligned}$

since, all the points on $\mathrm{L}$ are at a constant distance from the planes $\mathrm{P}_1$ and $\mathrm{P}_2$

$\therefore$ Line $\mathrm{L}$ is perpendicular to normal of plane $\mathrm{P}_1$ and $\mathrm{P}_2$

As we know if two lines are perpendicular then the sum of products of their respective direction ratios is zero.

$\begin{aligned} & \Rightarrow l+2 m-n=0 \ldots \ldots . \text { (iv) } \quad \left\{\because L \perp P_1\right\}\\ & \Rightarrow 2 l-m+n=0 \ldots \ldots . \text { (v) } \quad \left\{\because L \perp P_2\right\} \end{aligned}$

On solving equation (iv) and equation (v), we get

$\frac{l}{1}=\frac{m}{-3}=\frac{n}{-5}$

$\therefore \quad$ Equation of line L is $\frac{x}{1}=\frac{y}{-3}=\frac{z}{-5}$

Let any point on L is $\mathrm{A}(k,-3 k,-5 k)$

Now foot of perpendicular from $A$ to plane $P_1$ is given by

$\begin{aligned} & \frac{x-k}{1}=\frac{y+3 k}{2}=\frac{z+5 k}{-1}=-\frac{(k-6 k+5 k+1)}{1^2+2^2+1^2} \\ \Rightarrow \quad & \frac{x-k}{1}=\frac{y+3 k}{2}=\frac{z+5 k}{-1}=-\frac{1}{6} \\ \Rightarrow \quad & x=k-\frac{1}{6}, y=-3 k-\frac{1}{3}, z=-5 k+\frac{1}{6} \end{aligned}$

So, coordinates of foot of perpendicular is $\left(k-\frac{1}{6},-3 k-\frac{1}{3},-5 k+\frac{1}{6}\right)$

For $k=0$, foot of perpendicular $\left(-\frac{1}{6},-\frac{1}{3}, \frac{1}{6}\right)$

For $k=\frac{1}{6}$, foot of perpendicular $\left(0,-\frac{5}{6},-\frac{2}{3}\right)$

Hint :

(i) since all points on L are at constant distance from the planes $\mathrm{P}_1$ and $\mathrm{P}_2$. So line $L$ is perpendicular to normal of $P_1$ and $P_2$.

(ii) Find equation of line L using above condition and assume any point on L .

(iii) Use coordinates of foot of perpendicular drawn from a point $\left(x_1 y_1 z_1\right)$ to the plane $a x+b y+c z+d=0$ is given by

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=-\frac{a x_1+b y_1+c z_1+d}{a^2+b^2+c^2}$

$\begin{aligned} & \text { Given, } \mathrm{P}_1: y=0 \quad \text{... (i)}\\ & \mathrm{P}_2: x+z-1=0 \quad \text{... (ii)} \end{aligned}$

Equation of plane $P_3$ passing through the intersection of plane $P_1$ and $P_2$ is given by

$\mathrm{P}_3: x+z-1+\lambda y=0 \quad \text{... (iii)}$

$\because$ Distance of the point $(0,1,0)$ from $P_3$ is 1.

$\begin{aligned} & \therefore \quad\left|\frac{0+\lambda+0-1}{\sqrt{1^2+\lambda^2+1^2}}\right|=1 \\ & \Rightarrow \quad\left|\frac{\lambda-1}{\sqrt{2+\lambda^2}}\right|=1 \\ & \Rightarrow \quad(\lambda-1)^2=1\left(\sqrt{2+\lambda^2}\right)^2 \\ & \Rightarrow \quad \lambda^2+1-2 \lambda=\lambda^2+2 \\ & \Rightarrow \quad \lambda=-\frac{1}{2} \\ \end{aligned}$

Put the value of $\lambda$ in equation (iii), we get

$\begin{aligned} & \mathrm{P}_3: \quad x+z-1-\frac{1}{2} y=0 \\ & \Rightarrow \quad 2 x-y+2 z-2=0 \end{aligned}$

$\because$ Distance of a point $(\alpha, \beta, \gamma)$ from $\mathrm{P}_3$ is 2

$\begin{array}{ll} \therefore & \left|\frac{2 \alpha-\beta+2 \gamma-2}{\sqrt{2^2+1^2+2^2}}\right|=2 \\ \Rightarrow & \left|\frac{2 \alpha-\beta+2 \gamma-2}{3}\right|=2 \\ \Rightarrow & 2 \alpha-\beta+2 \gamma-2= \pm 6 \\ \Rightarrow & 2 \alpha-\beta+2 \gamma=8 \text { or }-4 \\ \Rightarrow & 2 \alpha-\beta+2 \gamma-8=0 \text { or } 2 \alpha-\beta+2 \gamma+4=0 \end{array}$

Hint :

(i) Equation of plane passing through the intersection of two plane $P_1$ and $P_2$ is given by $\mathrm{P}_1+\lambda \mathrm{P}_2=0$; where $\lambda$ is a constant.

(ii) Distance of a point $\left(x_1 y_1 z_1\right)$ from plane $a x +b y+c z+d=0$ is $\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

${{a - 1} \over 2} = {{b - 3} \over { - 1}} = {{c - 4} \over 1} = \lambda \left( {let} \right)$

$ \Rightarrow a = 2\lambda + 1,\,\,b = 3 - \lambda ,\,\,c = 4 + \lambda $

$p = \left( {{{a + 1} \over 2},{{b + 3} \over 2},{{c + 4} \over 2}} \right)$

$ = \left( {\lambda + 1,{{6 - \lambda } \over 2},{{\lambda + 8} \over 2}} \right)$

$\therefore$ $2\left( {\lambda + 1} \right) - {{6 - \lambda } \over 2} + {{\lambda + 8} \over 2} + 3 = 0$

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

$a = - 3,b = 5,c = 2$

Required line is ${{x + 3} \over 3} = {{y - 5} \over 1} = {{z - 2} \over { - 5}}$

Given

$l + m + n = 0$ and ${l^2} = {m^2} + {n^2}$

Now, ${\left( { - m - n} \right)^2} = {m^2} + {n^2}$

$ \Rightarrow mn = 0 \Rightarrow m = 0\,\,$ or $\,\,n = 0$

If $m=0$ then $l=-n$

We know

${l^2} + {m^2} + {n^2} = 1 \Rightarrow n = \pm {1 \over {\sqrt 2 }}$

i.e.$\left( {{l_1},{m_1},{n_1}} \right) = \left( { - {1 \over {\sqrt 2 }},0,{1 \over {\sqrt 2 }}} \right)$

If $n=0$ then $l=-m$

${l^2} + {m^2} + {n^2} = 1\,\,\, \Rightarrow 2{m^2} = 1$

$ \Rightarrow m = \pm {1 \over {\sqrt 2 }}$

Let $m = {1 \over {\sqrt 2 }} \Rightarrow l = - {1 \over {\sqrt 2 }}$

and $n=0$

$\left( {{l_2},{m_2},{n_2}} \right) = \left( { - {1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }},0} \right)$

$\therefore$ $\cos \theta = {1 \over 2} \Rightarrow \theta = {\pi \over 3}$

Line $L_1$ given by $y=x ; z=1$ can be expressed

$ L_1: \frac{X}{1}=\frac{Y}{1}=\frac{z-1}{0} $

$ \begin{aligned} \frac{x}{1} & =\frac{y}{1}=\frac{z-1}{0}=\alpha (say)\\\\ \Rightarrow x & =\alpha, y=\alpha, z=1 \end{aligned} $

Let the coordinates of $Q$ on $L_1$ be $(\alpha, \alpha, 1)$

Line $L_2$ given by $y=-x, z=-1$ can be expressed as

$L_2: \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}$

$\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=\beta $ (say)

$ \Rightarrow \quad x=\beta, y=-\beta, z=-1 $

Let the coordinates of $R$ on $L_2$ be $(\beta,-\beta,-1)$.

Direction ratios of $P Q$ are $\lambda-\alpha, \lambda-\alpha, \lambda-1$.

Now, $P Q \perp L_1$

$\begin{aligned} & \therefore 1(\lambda-\alpha)+1 \cdot(\lambda-\alpha)+0 \cdot(\lambda-1)=0 \\\\ & \Rightarrow \lambda=\alpha \\\\ & \therefore Q(\lambda, \lambda, 1)\end{aligned}$

Direction ratio of $P R$ are

$ \lambda-\beta, \lambda+\beta, \lambda+1 \text {. } $

Now, $P R \perp L_2$

$\begin{aligned} & \therefore 1(\lambda-\beta)+(-1)(\lambda+\beta)+0(\lambda+1)=0 \\\\ & \lambda-\beta-\lambda-\beta=0 \\\\ & \Rightarrow \beta=0 \\ & \end{aligned}$
$\therefore R(0,0,-1)$

Now, as $ \angle Q P R=90^{\circ}$

$ \text { (as } a_1 a_2+b_1 b_2+c_1 c_2=0 \text {, } $ if two lines with DR's $a_1, b_1, c_1 ; a_2, b_2, c_2$ are perpendicular)

$\begin{array}{cc}\therefore & (\lambda-\lambda)(\lambda-0)+(\lambda-\lambda)(\lambda-0) +(\lambda-1)(\lambda+1)=0 \\\\ &\Rightarrow (\lambda-1)(\lambda+1)=0 \\\\ &\Rightarrow \lambda=1 \text { or } \lambda=-1\end{array}$

$\lambda=1$, rejected as $P$ and $Q$ are different points.

$ \Rightarrow \lambda=-1 $

Given lines will be coplanar

If $\,\,\,\,\left| {\matrix{ { - 1} & 1 & 1 \cr 1 & 1 & { - k} \cr k & 2 & 1 \cr } } \right| = 0$

$ \Rightarrow - 1\left( {1 + 2k} \right) - \left( {1 + {k^2}} \right) + 1\left( {2 - k} \right) = 0$

$ \Rightarrow k = 0, - 3$

$2x + y + 2z - 8 = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( {Plane\,\,1} \right)$

$2x + y + 2z + {5 \over 2} = 0\,\,\,\,\,\,\,\,\,\,\,...\left( {Plane\,\,2} \right)$

Distance between Plane $1$ and $2$

$ = \left| {{{ - 8 - {5 \over 2}} \over {\sqrt {{2^2} + {1^2} + {2^2}} }}} \right|$

$ = \left| {{{ - 21} \over 6}} \right| = {7 \over 2}$

Let $Q$ is the point of intersection of lines

$\begin{aligned} & \mathrm{L}_1: \frac{x-1}{2}=\frac{y}{-1}=\frac{z+3}{1}=\lambda \text { and } \\ & \mathrm{L}_2: \frac{x-4}{1}=\frac{y+3}{1}=\frac{z+3}{2}=\mu \\ & \mathrm{Q}=(1+2 \lambda,-\lambda,-3+\lambda) \\ & =(4+\mu,-3+\mu,-3+2 \mu) \\ & \Rightarrow \lambda=3-\mu \text { and } \lambda=2 \mu \\ & \Rightarrow \lambda=2 \mu=3-\mu \\ & \Rightarrow \mu=1, \lambda=2 \\ & \therefore \quad Q=(5,-2,-1) \\ \end{aligned}$

Given, the plane $a x+b y+c z=d$ passing through the intersection of lines $\mathrm{L}_1$ and $\mathrm{L}_2$

$\therefore \quad \mathrm{Q}(5,-2,-1)$ satisfy the equation

$\begin{aligned} & a x+b y+c z=d \\ \Rightarrow & 5 a-2 b-c=d \quad \text{... (i)} \end{aligned}$

Given, the plane $a x+b y+c z=d$ is perpendicular to both the planes $3 x+5 y-6 z =4$ and $7 x+y+2 z=3$

$\therefore \quad a \hat{i}+b \hat{j}+c \hat{k}$ is parallel to

$\begin{aligned} & (7 \hat{i}+\hat{j}+2 \hat{k}) \times(3 \hat{i}+5 \hat{j}-6 \hat{k}) \\ \Rightarrow & a \hat{i}+b \hat{j}+c \hat{k}=\gamma\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 7 & 1 & 2 \\ 3 & 5 & -6 \end{array}\right| \\ \Rightarrow & a \hat{i}+b \hat{j}+c \hat{k}=-16 \gamma \hat{i}+48 \gamma \hat{j}+32 \gamma \hat{k} \\ \Rightarrow & a=-16 \gamma, b=48 \gamma, c=32 \gamma \end{aligned}$

If $\gamma=\frac{-1}{16}$

$\Rightarrow a=1, b=-3, c=-2$

Put $a=1, b=-3$ and $c=-2$ in the equation (i)

$\begin{aligned} & \Rightarrow d=5+6+2 \\ & \Rightarrow d=13 \end{aligned}$

Hints:

(i) If a plane $\mathrm{P}=0$ is perpendicular to both planes $P_1=0$ and $P_2=0$, then normal vector of $P$ is parallel to cross Product of normals of planes $\mathrm{P}_1$ and $\mathrm{P}_2$.

(ii) Recall the method of finding the point of intersection of two lines in three Dimensional co-ordinate geometry.

Let $ P $ be a general point on the line :

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

Therefore, the coordinates of $ P $ can be expressed as :

$ P = (-2 + 2\lambda, -1 - \lambda, 3\lambda) $

Now, let $(h, k, w)$ represent the foot of the perpendicular from $ P $ to the plane $ x + y + z = 3 $.

We use the formula for the foot of the perpendicular from a point to a plane :

$ h = \frac{-1 \cdot (-2 + 2\lambda - 1 - \lambda + 3\lambda - 3)}{1^2 + 1^2 + 1^2} + (-2 + 2\lambda), \\ k = \frac{-1 \cdot (-2 + 2\lambda - 1 - \lambda + 3\lambda - 3)}{1^2 + 1^2 + 1^2} + (-1 - \lambda), \\ w = \frac{-1 \cdot (-2 + 2\lambda - 1 - \lambda + 3\lambda - 3)}{1^2 + 1^2 + 1^2} + 3\lambda $

Simplifying these, we get :

$ h = \frac{+2\lambda}{3}, \\ k = \frac{-7\lambda}{3} + 1, \\ w = \frac{5\lambda}{3} + 2 $

So, we can express $ \lambda $ and the coordinates $ (h, k, w) $ as :

$ \frac{h}{2} = \frac{k - 1}{-7} = \frac{w - 2}{5} = \frac{\lambda}{3} $

Therefore, the locus of the foot of the perpendiculars is :

$ \frac{x}{2} = \frac{y - 1}{-7} = \frac{z - 2}{5} = \frac{\lambda}{3} $

Thus, the line on which the feet of the perpendiculars lie is represented as :

$ \frac{x}{2} = \frac{y - 1}{-7} = \frac{z - 2}{5} $

Given two line

$\mathrm{L}_1: \frac{x-5}{0}=\frac{y}{3-\alpha}=\frac{z}{-2}$ and

$\mathrm{L}_2: \frac{x-\alpha}{0}=\frac{y}{-1}=\frac{z}{2-\alpha}$

$\Rightarrow \overrightarrow{\mathrm{AB}}=(\alpha-5) \hat{i}+0 \hat{j}+0 \hat{k}$

$\therefore \quad \mathrm{L}_1$ and $\mathrm{L}_2$ are coplanar

$\therefore$ The scalar triple product of $\overrightarrow{\mathrm{AB}}, 0 \hat{i}+(3-\alpha) \hat{j}-2 \hat{k}$ and $0 \hat{i}-\hat{j}+(2-\alpha) \hat{k}$ is zero.

$\begin{aligned} & \\ \Rightarrow \quad & \left|\begin{array}{ccc} \alpha-5 & 0 & 0 \\ 0 & 3-\alpha & -2 \\ \alpha & -1 & (2-\alpha) \end{array}\right|=0 \\ \Rightarrow \quad & (\alpha-5)[(3-\alpha)(2-\alpha)-2]=0 \\ \Rightarrow \quad & (\alpha-5)\left(\alpha^2-5 \alpha+4\right)=0 \\ \Rightarrow \quad & (\alpha-5)(\alpha-1)(\alpha-4)=0 \\ \Rightarrow \quad & \alpha=1,4,5 \end{aligned}$

Hints :

If two line $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ are coplanar, then $\left|\begin{array}{ccc}x_1-x_2 & y_1-y_2 & y_2-y_3 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{array}\right|=0$

Given two lines.

$\begin{gathered} l_1:(3+t) \hat{i}+(-1+2 t) \hat{j}+(4+2 t) \hat{k} \text { and } \\ l_2:(3+2 s) \hat{i}+(3+2 s) \hat{j}+(2+s) \hat{k} \\ \Rightarrow l_1: \frac{x-3}{1}=\frac{y+1}{2}=\frac{z-4}{2}=t \text { and } \\ \quad l_2: \frac{x-3}{2}=\frac{y-3}{2}=\frac{z-2}{1}=s \end{gathered}$

Given, a line $l$ is perpendicular to $l_1$ and $l_2$

$\therefore \quad l$ is parallel to $l_1 \times l_2$

$\begin{aligned} & \Rightarrow \vec{l}_1 \times \vec{l}_2=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{array}\right| \\ & \Rightarrow \vec{l}_1 \times \vec{l}_2=-2 \hat{i}+3 \hat{j}-2 \hat{k} \end{aligned}$

The equation of line $l$ passing through origin and parallel to $-2 \hat{i}+3 \hat{j}-2 \hat{k}$ is

$l: \frac{x}{-2}=\frac{y}{3}=\frac{z}{-2}=\lambda$

Let P is the point of intersection of $l$ and $l_1$

$\begin{aligned} & \Rightarrow \mathrm{P}=(-2 \lambda, 3 \lambda,-2 \lambda)=(3+t,-1+2 t, 4+2 t) \\ & \Rightarrow \lambda=\frac{3+t}{-2}=\frac{-1+2 t}{3}=\frac{4+2 t}{-2} \\ & \Rightarrow t=-1, \lambda=-1 \end{aligned}$

$\text { So, } \mathrm{P}=(2,-3,2)$

Let Q is a point on the line $l_2$

$Q=(3+2 s, 3+2 s, 2+s)$

Apply $|\mathrm{PQ}|=\sqrt{17}$

$\begin{aligned} & \Rightarrow \sqrt{(3+2 s-2)^2+(3+2 s+3)^2+(2+s-2)^2} \\ & =\sqrt{17} \\ & \Rightarrow \sqrt{9 s^2+28 s+37}=\sqrt{17} \\ & \Rightarrow \quad 9 s^2+28 s+37=17 \\ & \Rightarrow \quad 9 s^2+28 s+20=0 \\ & \Rightarrow \mathrm{s}=\frac{-28 \pm \sqrt{28^2-4 \times 9 \times 20}}{2.9} \\ & \Rightarrow \mathrm{s}=\frac{-28 \pm 8}{18} \\ & \Rightarrow \mathrm{s}=\frac{-10}{9},-2 \\ \end{aligned}$

If $s=-2$, then $Q=(-1,-1,0)$

$\text { If } s=-\frac{10}{9} \text {, then } Q=\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$

Hints :

(i) If $l$ is perpendicular to $\vec{l}_1$ and $\vec{l}_2$, then $l$ is parallel to $\vec{l}_1 \times \vec{l}_2$.

(ii) A general point on the line

$\begin{gathered} \quad \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\lambda \text { is } \\ \left(x_1+\lambda a, y_1+\lambda b, z_1+\lambda c\right) . \end{gathered}$

Given lines in vector form are

$\overrightarrow r = \left( {\widehat i - \overrightarrow j + \overrightarrow k } \right) + \lambda \left( {2\overrightarrow i + 3\overrightarrow j + 4\overrightarrow j } \right)$

and $\overrightarrow r = \left( {3\widehat i + k\widehat j} \right) + \mu \left( {\widehat i + 2\widehat j + \widehat k} \right)$

These will intersect if shortest distance between them $=0$

i.e.$\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).\overrightarrow {{b_1}} \times {\overrightarrow b _2} = 0$

$ \Rightarrow \left| {\matrix{ {3 - 1} & {k + 1} & { - 1} \cr 2 & 3 & 4 \cr 1 & 2 & 1 \cr } } \right| = 0$

$ \Rightarrow 2\left( { - 5} \right) - \left( {k + 1} \right)\left( { - 2} \right) - 1\left( 1 \right) = 0$

$ \Rightarrow k = 9/2$

Given equation of a plane is

$x - 2y + 2z - 5 = 0$

So, Equation of parallel plane is given by

$x - 2y + 2z + d = 0$

Now, it is given that distance from origin to the parallel planes is $1.$

$\therefore$ $\left| {{d \over {\sqrt {{1^2} + {2^2} + {2^3}} }}} \right| = 1 \Rightarrow d = \pm 3$

So equation of required plane

$x - 2y + 2z \pm 3 = 0$

The equation of plane passing through the intersection of planes $x+2 y+3 z-2=0$ and $x-y+z-3=0$ is

$\begin{aligned} & \Rightarrow(x+2 y+3 z-2)+\lambda(x-y+z-3)=0 \\ & \Rightarrow(\lambda+1) x+(-\lambda+2) y+(\lambda+3) z-(3 \lambda+2)=0 \quad \text{... (i)} \end{aligned}$

$\text { Given, the distance from }(3,1,-1) \text { is } \frac{2}{\sqrt{3}}$

$\begin{aligned} & \Rightarrow \frac{2}{\sqrt{3}}=\frac{|3(\lambda+1)+1 \cdot(-\lambda+2)-1 \cdot(\lambda+3)-(3 \lambda+2)|}{\sqrt{(\lambda+1)^2+(-\lambda+2)^2+(\lambda+3)^2}} \\ & \Rightarrow \frac{2}{\sqrt{3}}=\frac{|-2 \lambda|}{\sqrt{3 \lambda^2+4 \lambda+14}} \\ & \Rightarrow 2 \sqrt{3 \lambda^2+4 \lambda+14}=2|\lambda| \sqrt{3} \\ & \Rightarrow \sqrt{3 x^2+4 \lambda+14}=\sqrt{3}|\lambda| \end{aligned}$

On squaring both side

$\begin{aligned} & \Rightarrow 3 \lambda^2+4 \lambda+14=3 \lambda^2 \\ & \Rightarrow \lambda=-\frac{7}{2} \end{aligned}$

Put $\lambda=-\frac{7}{2}$ in the equation (i)

$\begin{aligned} & \Rightarrow\left(-\frac{7}{2}+1\right) x+\left(\frac{7}{2}+2\right) y+\left(-\frac{7}{2}+3\right) z +\frac{21}{2}-2=0 \\ & \Rightarrow \frac{-5}{2} x+\frac{11}{2} y-\frac{1}{2} z+\frac{17}{2}=0 \\ & \Rightarrow \quad 5 x-11 y+z=17 \\ \end{aligned}$

Equation of line QR is

$\frac{x-2}{1}=\frac{y-3}{4}=\frac{z-5}{1}=\lambda$

General point on QR will be $(\lambda+2,4 \lambda+3, \lambda+5)$

Let $P \equiv(\lambda+2,4 \lambda+3, \lambda+5)$

Since, point P also lies on the plane.

$\begin{array}{rrrl} \therefore & 5(\lambda+2)-4(4 \lambda+3)-(\lambda+5) & =1 \\ \Rightarrow & 5 \lambda+10-16 \lambda-12-\lambda-5 & =1 \\ \Rightarrow & -12 \lambda-7 & =1 \end{array}$

$\begin{array}{rrrl} \Rightarrow & \lambda &=\frac{-8}{12} \\ \Rightarrow & \lambda &=\frac{-2}{3} \end{array}$

Hence,

$\begin{aligned} P & \equiv\left(\frac{-2}{3}+2,4\left(\frac{-2}{3}\right)+3, \frac{-2}{3}+5\right) \\ & \equiv\left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right) \end{aligned}$

Now, for TS to be perpendicular to QR,

$ \begin{array}{rlrl} & \lambda+4(4 \lambda+2)+(\lambda+1) & =0 \\ \Rightarrow & \lambda \lambda+16 \lambda+8+\lambda+1 & =0 \\ \Rightarrow & 18 \lambda =-9 \\ \Rightarrow & \lambda =\frac{-1}{2} \end{array}$

$\begin{aligned} \therefore \quad \text { Point } S & \equiv\left(\frac{-1}{2}+2,4\left(\frac{-1}{2}\right)+3, \frac{-1}{2}+5\right) \\ & \equiv\left(\frac{3}{2}, 1, \frac{9}{2}\right) \end{aligned}$

$\therefore$ Length of line segment PS

$\begin{aligned} & =\sqrt{\left(\frac{4}{3}-\frac{3}{2}\right)^2+\left(\frac{1}{3}-1\right)^2+\left(\frac{13}{3}-\frac{9}{2}\right)^2} \\ & =\sqrt{\frac{1}{36}+\frac{4}{9}+\frac{1}{36}} \\ & =\sqrt{\frac{1+16+1}{36}} \\ & =\sqrt{\frac{18}{36}} \\ & =\frac{1}{\sqrt{2}} \end{aligned}$

Given, the lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar.

Apply the scalar triple product of $(2 \hat{i}+k \hat{j}+2 \hat{k}),(5 \hat{i}+2 \hat{j}+k \hat{k})$ and $(-2 \hat{i}+0 \hat{j}+0 \hat{k})$ is zero.

$\begin{aligned} & \Rightarrow\left|\begin{array}{ccc} -2 & 0 & 0 \\ 2 & k & 2 \\ 5 & 2 & k \end{array}\right|=0 \\ & \Rightarrow-2\left(k^2-4\right)=0 \\ & \Rightarrow \quad k= \pm 2 \\ \end{aligned}$

For the equation of plane containing given lines, apply scalar triple product of $(x-1) \hat{i}+(y+1) \hat{j}+z \hat{k}, 2 \hat{i}$ and $2 \hat{i}+k \hat{j}+2 \hat{k}$ is zero.

$\begin{array}{lrl} \Rightarrow & \left|\begin{array}{ccc} x-1 & y+1 & z \\ -2 & 0 & 0 \\ 2 & \pm 2 & 2 \end{array}\right|=0 \\ \Rightarrow & 4(y+1)-2 k z=0 \\ \Rightarrow & 2(y+1)-( \pm 2) z=0 \\ \Rightarrow & y+1 \mp z=0 \\ \Rightarrow & y-z+1=0, y+z+1=0 \end{array}$

If $\theta $ be the angle between the given line and plane, then

$\sin \theta = {{1 \times 1 + 2 \times 2 + \lambda \times 3} \over {\sqrt {{1^2} + {2^2} + {\lambda ^2}} .\sqrt {{1^2} + {2^2} + {3^2}} }}$

$ = {{5 + 3\lambda } \over {\sqrt {14} .\sqrt {5 + {\lambda ^2}} }}$

But it is given that

$\theta = {\cos ^{ - 1}}\sqrt {{5 \over {14}}} \Rightarrow \sin \theta = {3 \over {\sqrt {14} }}$

$\therefore$ ${{5 + 3\lambda } \over {\sqrt {14} \sqrt {5 + {\lambda ^2}} }} = {3 \over {\sqrt {14} }} \Rightarrow \lambda = {2 \over 3}$

The directions ratios of the line segment joining points

$A\left( {1,0,7} \right)\,\,$ and $\,\,\,B\left( {1,6,3} \right)$ are $0,6, - 4.$

The direction ratios of the given line are $1,2,3.$

Clearly $1 \times 0 + 2 \times 6 + 3 \times \left( { - 4} \right) = 0$

So, the given line is perpendicular to line $AB.$

Also, the mid point of $A$ and $B$ is $\left( {1,3,5} \right)$ which lies on the given line.

So, the image of $B$ in the given line is $A$, because the given line is the perpendicular bisector of line segment joining points $A$ and $B$, But statement -$2$ is not a correct explanation for statement$-1.$

Direction cosines of the line :

$\ell = \cos {45^ \circ } = {1 \over {\sqrt 2 }},m = \cos {120^ \circ } = {{ - 1} \over 2},\pi = \cos \theta $

where $\theta $ is the angle, which line makes with positive $z$-axis.

Now ${\ell ^2} + {m^2} + {n^2} = 1$

$ \Rightarrow {1 \over 2} + {1 \over 4} + {\cos ^2}\theta = 1,\,\,{\cos ^2}\theta = {1 \over 4}$

$ \Rightarrow \cos \theta = {1 \over 2}\,\,\,\,\left( \theta \right.$ being acute)

$ \Rightarrow 0 = {\pi \over 3}$

$A\left( {3,1,6} \right);\,\,B = \left( {1,3,4} \right)$

Mid-point of $AB = \left( {2,2,5} \right)$ lies on the plane.

and d.r's of $AB=(2,-2,2)$

d.r's of normal to plane $ = \left( {1, - 1,1} \right).$

Direction ratio of $AB$ and normal to the plane are proportional therefore,

$AB$ is perpendicular to the plane

$\therefore$ $A$ is image of $B$

Statement-$2$ is correct but it is not correct explanation.

Plane 1 : $ax + by + cz = 0$ contains line ${x \over 2} = {y \over 3} = {z \over 4}$. Therefore,

$2a + 3b + 4c = 0$ ..... (1)

Plane 2 : $a'x + b'y + c'z = 0$ is perpendicular to plane containing line ${x \over 3} = {y \over 4} = {z \over 2}$ and ${x \over 4} = {y \over 2} = {z \over 3}$.

Hence, $3a' + 4b' + 2c' = 0$ and $4a' + 2b' + 3c' = 0$.

$ \Rightarrow {{a'} \over {12 - 4}} = {{b'} \over {8 - 9}} = {{c'} \over {6 - 16}}$

$ \Rightarrow 8a - b - 10c = 0$ .... (2)

From Eq. (1) and (2), we get

${a \over { - 30 + 4}} = {b \over {32 + 20}} = {c \over { - 2 - 24}}$

$\Rightarrow$ Equation of plane is $x - 2y + z = 0$

(A) Equation of the line passing through origin is

${x \over a} = {y \over b} = {z \over c}$

$\therefore$ $\left| {\matrix{ 2 & 1 & { - 1} \cr 1 & { - 2} & 1 \cr a & b & c \cr } } \right| = 0$

$ \Rightarrow a( - 1) - b(3) + c( - 5) = 0$

$ \Rightarrow - a - 3b - 5c = 0$

$ \Rightarrow a + 3b + 5c = 0$ ....... (i)

Also, $\left| {\matrix{ {{8 \over 3}} & { - 3} & 1 \cr 2 & { - 1} & 1 \cr a & b & c \cr } } \right| = 0$

$\therefore$ $a( - 2) - b\left( {{2 \over 3}} \right) + c\left( {{{10} \over 3}} \right) = 0$

$ \Rightarrow 2a + {{2b} \over 3} - {{10c} \over 3} = 0$

$3a + b - 5c = 0$ ...... (ii)

From Eqs. (i) and (ii),

${a \over { - 20}} = {b \over {20}} = {c \over { - 8}}$

${a \over 5} = {b \over { - 5}} = {c \over 4}$

Equation of line is

${x \over 5} = {y \over { - 5}} = {z \over 4} = \lambda $ (say) ....... (iii)

Also, ${{x - 2} \over 1} = {{y - 1} \over 2} = {{z + 1} \over 1} = {k_1}$ (say) ...... (iv)

Now, ${{x - {8 \over 3}} \over 2} = {{y + 3} \over { - 1}} = {{z - 1} \over 1} = {k_2}$ (say) ...... (v)

Point on (iii) is $(5\lambda ,\, - 5\lambda ,\, + 4\lambda )$

Point on (iv) is $(2 + {k_1},\,1 - 2{k_1},\, - 1 + {k_1})$

Point on (v) is $\left( {{8 \over 3} + 2{k_2},\, - 3 - {k_2},\,1 + {k_2}} \right)$

On solving, $2 + {k_1} + 1 - 2{k_1} = 0$

$ - {k_1} + 3 = 0$

${k_1} = 3$

$P \equiv (5,\, - 5,\,2)$

Again, for Q

${8 \over 3} + 2{k_2} - 3 - {k_2} = 0$

${k_2} - {1 \over 3} = 0$

${k_2} = {1 \over 3}$

$Q \equiv \left( {{{10} \over 3},{{ - 10} \over 3},{4 \over 3}} \right)$

Now, $PQ = \sqrt {{{\left( {{5 \over 3}} \right)}^2} + {{\left( {{5 \over 3}} \right)}^2} + {{\left( {{2 \over 3}} \right)}^2}} $

$ = {{\sqrt {54} } \over 3}$

$P{Q^2} = {d^2} = {{54} \over 9} = 6$

(B) ${\tan ^{ - 1}}\left( {{{x + 3 - x + 3} \over {1 + ({x^2} - 9)}}} \right) = {\tan ^{ - 1}}\left( {{3 \over 4}} \right)$

$ \Rightarrow {6 \over {{x^2} - 8}} = {3 \over 4}$

$ \Rightarrow 3{x^2} = 48$

$ \Rightarrow x = \, \pm \,4$

(C) $\left( {\overrightarrow b - \overrightarrow a } \right)\,.\,\left( {\overrightarrow b + {{\overrightarrow a - \mu \overrightarrow b } \over 4}} \right) = 0$

$ \Rightarrow \left( {\overrightarrow b - \overrightarrow a } \right)\,.\,\left( {4\overrightarrow b + \overrightarrow a - \mu \overrightarrow b } \right) = 0$

$\left( {4 - \mu } \right){\overrightarrow b ^2} - {\overrightarrow a ^2} = 0$ ...... (i)

Also, $2\left| {\overrightarrow b + {{\overrightarrow a - \mu \overrightarrow b } \over 4}} \right| = \left| {\overrightarrow b - \overrightarrow a } \right|$

$ \Rightarrow 2\left| {{{(4 - \mu )\overrightarrow b + \overrightarrow a } \over 4}} \right| = \left| {\overrightarrow b - \overrightarrow a } \right|$

Distance of point P from plane = 5

$\therefore$ $5 = \left| {{{1 - 4 - 2 - \alpha } \over 3}} \right|$

$\alpha = 10$

Foot of perpendicular

${{x - 1} \over 1} = {{y + 2} \over 2} = {{z - 1} \over { - 2}} = {5 \over 3}$

$ \Rightarrow x = {8 \over 3},y = {4 \over 3},z = - {7 \over 3}$

Thus, the foot of the perpendicular is

$A\left( {{8 \over 3},{4 \over 3}, - {7 \over 3}} \right)$

As the line ${{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2}$ lie in the plane

$x + 3y - \alpha z + \beta = 0$

$\therefore$ $Pt\left( {2,1, - 2} \right)$ lies on the plane

i.e. $2 + 3 + 2\alpha + \beta = 0$

or $\,\,\,\,2\alpha + \beta + 5 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$

Also normal to plane will be perpendicular to line,

$\therefore$ $\,\,\,3 \times 1 - 5 \times 3 + 2 \times \left( { - \alpha } \right) = 0$

$ \Rightarrow \alpha = - 6$

From equation $(i)$ then, $\beta = 7$

$\therefore$ $\left( {\alpha ,\beta } \right) = \left( { - 6,7} \right)$

Let $P\left( {{x_1},{y_1},{z_1}} \right)$ and $Q\left( {{x_2},{y_2},{z_2}} \right)$ be the initial and final points of the vector whose projections on the three coordinates axes are ${6, - 3,2}$ then

${x_2} - {x_1}, = 6;\,\,{y_2} - {y_1} = - 3;\,\,{z_2} - {z_1} = 2$

So that directions ratios of $\overrightarrow {PQ} $ are ${6, - 3,2}$

$\therefore$ Direction cosines of $\overrightarrow {PQ} $ are

${6 \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }},{{ - 3} \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }},$

$\,\,\,\,\,\,\,\,$ ${2 \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }} = {6 \over 7},{{ - 3} \over 7},{2 \over 7}$

The D.C. of the line are ${1 \over {\sqrt 3 }},{1 \over {\sqrt 3 }},{1 \over {\sqrt 3 }}$.

We find that any point on the line at a distance $t$ from $P(2, - 1,2)$ is

$\left( {2 + {t \over {\sqrt 3 }}, - 1 + {t \over {\sqrt 3 }},2 + {t \over {\sqrt 3 }}} \right)$ which lies on $2x + y + z = 9 $

$\Rightarrow t = \sqrt 3 $.

We see that any point on the line is given as

$Q \equiv \{ (1 - 3\mu ),(\mu - 1),(5\mu + 2)\} $

$\overrightarrow {PQ} \equiv \{ - 3\mu - 2,\mu - 3,5\mu - 4\} $

Now, we have

$1( - 3\mu - 2) - 4(\mu - 3) + 3(5\mu - 4) = 0$

$ \Rightarrow - 3\mu - 2 - 4\mu + 12 + 15\mu - 12 = 0$

That is, $8\mu = 2 \Rightarrow \mu = {1 \over 4}$

Equation of line through $\left( {5,1,a} \right)$ and

$\left( {3,b,1} \right)$ is ${{x - 5} \over { - 2}} = {{y - 1} \over {b - 1}} = {{z - a} \over {1 - a}} = \lambda $

$\therefore$ Any point on this line is a

$\left[ { - 2\lambda + 5,\left( {b - 1} \right)\lambda + 1,\left( {1 - \alpha } \right)\lambda + a} \right]$

It crosses $yz$ plane where ${ - 2\lambda + 5 = 0}$

$\lambda = {5 \over 2}$

$\therefore$ $\left( {0,\left( {b - 1} \right){5 \over 2} + 1,\left( {1 - a} \right){5 \over 2} + a} \right) = \left( {0,{{17} \over 2},{{ - 17} \over 2}} \right)$

$ \Rightarrow \left( {b - 1} \right){5 \over 2} + 1 = {{17} \over 2}$

and $\left( {1 - a} \right){5 \over 2} + a = - {{13} \over 2}$

$ \Rightarrow b = 4$ and $a = 6$

The two lines intersect if shortest distance between them is zero $i.e.$

${{\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}} = 0$

$ \Rightarrow \left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{\overrightarrow b _1} \times {\overrightarrow b _2} = 0$

where ${\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat k,{\overrightarrow b _1} = k\widehat i + 2\widehat j + 3\widehat k$

$\,\,\,\,\,\,\,{\overrightarrow a _2} = 2\widehat i + 3\widehat j + \widehat k,\,\,{\widehat b_2} = 3\widehat i + k\widehat j + 2\widehat k$

$ \Rightarrow \left| {\matrix{ 1 & 1 & { - 2} \cr k & 2 & 3 \cr 3 & k & 2 \cr } } \right| = 0$

$ \Rightarrow 1\left( {4 - 3k} \right) - 1\left( {2k - 9} \right) - 2\left( {{k^2} - 6} \right) = 0$

$ \Rightarrow - 2{k^2} - 5k + 25 = 0 \Rightarrow k = - 5$ $\,\,\,\,$ or $\,\,\,\,$ ${5 \over 2}$

As $k$ is an integer, therefore $k=-5$

The equation of the plane passing through the point ($-1,-2,-1$) and whose normal is perpendicular to both the given lines L$_1$ and L$_2$ written as

$(x + 1) + 7(y + 2) - 5(z + 1) = 0$

i.e., $x + 7y - 5z + 10 = 0$

The distance of the point (1, 1, 1) from the plane

$ = \left| {{{1 + 7 - 5 + 10} \over {\sqrt {1 + 49 + 25} }}} \right|$

$ = {{13} \over {\sqrt {75} }}$ units

We have,

${P_1}:x - y + z = - 1$

${P_2}:x + y - z = - 1$

${P_3}:x - 3y + 3z = 2$

Let dr's of the lines of L$_1$, L$_2$ and L$_3$ are ${a_1},{b_1},{c_1}:{a_2},{b_2},{c_2}$ and ${a_3},{b_3},{c_3}$ respectively.

Therefore,

${a_1} + {b_1} - {c_1} = 0$

${a_1} - 3{b_1} + 3{c_1} = 0$

$ \Rightarrow {{{a_1}} \over 0} = {{{b_1}} \over { - 4}} = {{{c_1}} \over { - 4}}$

${a_1},{b_1},{c_1} = 0,1,1$

again ${a_2} - {b_2} + {c_2} = 0$

${a_2} - 3{b_2} + 3{c_2} = 0$

${{{a_2}} \over 0} = {{{b_2}} \over { - 2}} = {{{c_2}} \over { - 2}}$

${a_2},{b_2},{c_2} = 0,1,1$

Again ${a_3} - {b_3} + {c_3} = 0$

${a_3} + {b_3} - {c_3} = 0$

$ \Rightarrow {{{a_3}} \over 0} = {{{b_3}} \over 2} = {{{c_3}} \over 2} \Rightarrow {a_3},{b_3},{c_3} = 0,1,1$

L$_1$, L$_2$ and L$_3$ are parallel.

For given sphere center is $\left( {3,6,1} \right)$

Coordinates of one end of diameter of the sphere are $\left( {2,3,5} \right).$

Let the coordinates of the other end of diameter are $\left( {\alpha ,\beta ,\gamma } \right)$

$\therefore$ ${{\alpha + 2} \over 2} = 3,\,\,{{\beta + 3} \over 2} = 6,\,\,{{\gamma + 5} \over 2} = 1$

$ \Rightarrow \alpha = 4,\,\,\beta = 9\,\,$ and $\,\,\gamma = - 3$

$\therefore$ Coordinate of other end of diameter are $\left( {4,9, - 3} \right)$

Let the direction cosines of line $L$ be $l,m,n,$

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

and $l + 3m + 2n = 0\,\,\,\,\,\,\,\,\,\,....\left( {ii} \right)$

on solving equation $(i)$ and $(ii),$ we get

${l \over {6 - 3}} = {m \over {1 - 4}} = {n \over {6 - 3}}$

$\,\,\,\,\,\,\,\,\,\, \Rightarrow {l \over 3} = {m \over { - 3}} = {n \over 3}$

Now $ \Rightarrow {l \over 3} = {m \over { - 3}} = {n \over 3} = {{\sqrt {{l^2} + {m^2} + {n^2}} } \over {\sqrt {{3^2} + {{\left( { - 3} \right)}^2} + {3^2}} }}$

As ${l^2} + {m^2} + {n^2} = 1$

$\therefore$ ${l \over 3} = {m \over { - 3}} = {n \over 3} = {1 \over {\sqrt {27} }}$

$ \Rightarrow l = {3 \over {\sqrt {27} }} = {1 \over {\sqrt 3 }},\,\,m = - {1 \over {\sqrt 3 }},n = {1 \over {\sqrt 3 }}$

Line $L,$ makes an angle $\alpha $ with $+ve$ $x$-axis

$\therefore$ $l = \cos \,\alpha \,\,\,\, \Rightarrow \,\,\,\cos \alpha \,\, = {1 \over {\sqrt 3 }}$

Let the angle of line makes with the positive direction of $z$-axis is $\alpha $ direction cosines of line with the $+ve$ directions of $x$-axis, $y$-axis, and $z$-axis is $l,$ $m,$ $n$ respectively.

$\therefore$ $l = \cos {\pi \over 4},m = \cos {\pi \over 4},\,\,n = cos\,\alpha $

as we know that, ${l^2} + {m^2} + {n^2} = 1$

$\therefore$ ${\cos ^2}{\pi \over 4} + {\cos ^2}{\pi \over 4} + {\cos ^2}\alpha = 1$

$ \Rightarrow {1 \over 2} + {1 \over 2} + {\cos ^2}\alpha = 1$

$ \Rightarrow {\cos ^2}\alpha = 0 \Rightarrow \alpha = {\pi \over 2}$

Hence, angle with positive direction of the $z$-axis is ${\pi \over 2}$

The line of intersection of given plane is

$3 x-6 y-2 z-15=0$

$2 x+y-2 z=5 \Rightarrow 2 x+y-2 z-5=0$

For $z=0$, we obtain

$x=3$ and $y=-1$

$\therefore$ line passes through $(3,-1,0)$

Let $a, b, c$ be the d'rs of line of intersection, then $3 a-6 b-2 c=0$ and $2 a+b-2 c=0$

Solving the above equation using cross product method

$a: b: c=14: 2: 15$

$\therefore$ Equation of line is

$\frac{x-3}{14}=\frac{y+1}{2}=\frac{z}{15}=t$

Whose parametric form is $x=3+14 t, y=-1+2 t, z=15 t$

$\therefore$ Statement $\mathrm{I}$ is false

Since $\mathrm{dr}$'s of line intersection of given planes are $14,2,15$

$\therefore 14 \hat{i}+2 \hat{j}+15 \hat{k}$ is parallel to this line.

Statement 2 is true.

Equation of lines

${{x - b} \over a} = {y \over 1} = {{z - d} \over c}$

${{x - b'} \over {a'}} = {y \over 1} = {{z - d'} \over {c'}}$

Line are perpendicular

$ \Rightarrow aa' + 1 + cc' = 0$

$E{q^n}\,\,\,\,$ of $\,\,\,\,PN: $-

${{x + 1} \over 1} = {{y - 3} \over { - 2}} = {{z - 4} \over 0} = \lambda $

$N\left( {\lambda - 1, - 2\lambda + 3 - 4} \right)$

It lies on $x-2y=0$

$ \Rightarrow \lambda - 1 + 4\lambda - 6 = 0$

$ \Rightarrow \lambda = 7/5$

$N\left( {{2 \over 5},{1 \over 5},4} \right)$

$N$ is mid point of $PP'$

$\therefore$ $\alpha - 1 = {4 \over 5},\beta + 3 = {2 \over 5},r + 4 = 8$

$ \Rightarrow \alpha = {9 \over 5},\beta = {{ - 13} \over 5},r = 4$

$\therefore$ Image is $\left( {{9 \over 5},{{ - 13} \over 5},4} \right)$

(i) We need to find the value of $\tan t$ where $t = \sum\limits_{i=1}^\infty \tan^{-1}\left(\frac{1}{2i^2}\right)$.

We rewrite the sum: $\tan^{-1}\left(\frac{1}{2i^2}\right)$ can be written as $\tan^{-1}\left(\frac{2}{(2i+1)-(2i-1)}\right)$.

Using the formula: $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$, we turn the sum into a telescoping series: $\sum\limits_{i=1}^\infty [\tan^{-1}(2i+1) - \tan^{-1}(2i-1)]$.

When you write out the terms, most of them cancel out. What is left is: $t = \lim_{n \to \infty} [\tan^{-1}(2n+1) - \tan^{-1}(1)]$.

As $n$ becomes very large, $\tan^{-1}(2n+1)$ approaches $\frac{\pi}{2}$, so: $t = \frac{\pi}{2} - \tan^{-1}(1)$.

We know that $\tan^{-1}(1) = \frac{\pi}{4}$, so: $t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

So, $\tan t = \tan \left(\frac{\pi}{4}\right) = 1$.

(ii) The sides $a, b, c$ of a triangle are in arithmetic progression (AP). We have: $\cos \theta_1 = \frac{a}{b+c}, \cos \theta_2 = \frac{b}{a+c}, \cos \theta_3 = \frac{c}{a+b}$.

We know: $\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$.

So, $\cos \theta_1 = \frac{1-\tan^2\left(\frac{\theta_1}{2}\right)}{1+\tan^2\left(\frac{\theta_1}{2}\right)} = \frac{a}{b+c}$.

Using componendo and dividendo, we get: $\tan^2\left(\frac{\theta_1}{2}\right) = \frac{b+c-a}{a+b+c}$.

Similarly, $\tan^2\left(\frac{\theta_3}{2}\right) = \frac{a+b-c}{a+b+c}$.

So if we add them: $\tan^2\left(\frac{\theta_1}{2}\right) + \tan^2\left(\frac{\theta_3}{2}\right) = \frac{b+c-a + a+b-c}{a+b+c} = \frac{2b}{a+b+c}$.

If $a,b,c$ are in AP, then $a + c = 2b$, so $a + b + c = 3b$.

Therefore, $\tan^2\left(\frac{\theta_1}{2}\right) + \tan^2\left(\frac{\theta_3}{2}\right) = \frac{2b}{3b} = \frac{2}{3}$.

(iii) The line passes through (0, 1, 0) and is perpendicular to the plane $x + 2y + 2z = 0$.

The normal vector to the plane is (1, 2, 2). So, the direction vector for the line is (1, 2, 2).

The equation of the line is: $\frac{x-0}{1} = \frac{y-1}{2} = \frac{z-0}{2} = r$.

Let $P(r, 2r+1, 2r)$ be a point on the line. To find the value of $r$ where the vector from the origin (0, 0, 0) to the point on the line is perpendicular to the direction vector (1, 2, 2), set up:

$(r, 2r+1, 2r) \cdot (1, 2, 2) = 0$

That is: $r \times 1 + (2r+1)\times 2 + 2r \times 2 = 0$

So: $r + 4r + 2 + 4r = 0$ $9r + 2 = 0$ $r = -\frac{2}{9}$

So the point is: $\left(-\frac{2}{9}, \frac{5}{9}, -\frac{4}{9}\right)$

The distance from the origin to this point is: $\sqrt{\left(-\frac{2}{9}\right)^2 + \left(\frac{5}{9}\right)^2 + \left(-\frac{4}{9}\right)^2} = \sqrt{\frac{4+25+16}{81}} = \sqrt{\frac{45}{81}} = \frac{\sqrt{5}}{3}$

We know that equation of plane passes through ( $1,-2,1$ )

$ a(x-1)+b(y+2)+c(z+1)=0 $

above plane is $\perp^r$ on $2 x-2 y+z=0$

$ \therefore \quad 2 a-2 b+c=0 $

Similarly   $ a-b+2 c=0 $

(as it is $\perp^r$ on $x-y+2 z=0$ )

$ \frac{a}{1}=\frac{b}{1}=\frac{c}{0} $

$\therefore \quad$ equation of plane is $x+y+1=0$

Distance of the plane from the point ( $1,2,2$ )

$ \begin{aligned} & =\frac{1+2+1}{\sqrt{1^2+1^2}} \\ & =2 \sqrt{2} \end{aligned} $

Let vector AO be parallel to line of intersection of planes $P_1$ and $P_2$ through origin.

$\therefore$ Normal to plane $\mathrm{P}_1$ is

$ \vec{a}_1=(2 \hat{j}+3 \hat{k}) \times(4 \hat{j}-3 \hat{k})=-18 \hat{i} $

Normal to plane $P_2$ is

$ \begin{aligned} \vec{a}_2 & =(\hat{j}-\hat{k}) \times(3 \hat{i}+3 \hat{j}) \\ & =3 \hat{i}-3 \hat{j}-3 \hat{k} \end{aligned} $

$\therefore \quad \overrightarrow{\mathrm{OA}}$ is parallel to $\pm\left(\overrightarrow{n_1} \times \overrightarrow{n_2}\right)$

$ =54 \hat{j}-54 \hat{k} $

Angle between $54(\hat{j}-\hat{k})$ and $(2 \hat{i}+\hat{j}-2 \hat{k})$

is $\cos \theta= \pm\left(\frac{54+108}{3 \times 54-\sqrt{2}}\right)= \pm \frac{1}{\sqrt{2}}$

$ \theta=\frac{\pi}{4}, \frac{3 \pi}{4} $