3D Geometry

$\begin{aligned} & L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}=a \\ & L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}=b \end{aligned}$

$\begin{aligned} & x=a-11=3 b-16 \Rightarrow a-3 b=-5 \quad \text{.... (1)}\\ & y=2 a-21=2 b-11 \Rightarrow 2 a-2 b=10 \quad \text{.... (2)}\\ & z=3 a-29=b r-4 \Rightarrow 3 a-b y=25 \quad \text{.... (3)} \end{aligned}$

$\begin{aligned} & \text { from (1) & (2) } \\ & a=10, b=5 \end{aligned}$

Now from (3)

$\begin{aligned} & 3(10)-5 \gamma=25 \quad \therefore \gamma=1 \\ & \mathrm{R}_1 \equiv(-1,-1,1) \\ & \mathrm{OR}_1=-\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} \end{aligned}$

$\begin{aligned} & \vec{n}=\left|\begin{array}{lll} i & j & k \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right|=-4 \hat{i}-(-8) \hat{j}-4 \hat{k} \\ & \vec{n}=-4 \hat{i}+8 \hat{j}+4 \hat{k}=-4(\hat{i}-2 \hat{j}+\hat{k}) \\ & \hat{n}= \pm \frac{4(\hat{i}-2 \hat{j}+\hat{k})}{4 \sqrt{6}}= \pm \frac{(\hat{i}-2 \hat{j}+\hat{k})}{\sqrt{6}} \\ & \overrightarrow{O R} \cdot \hat{n}= \pm(-\hat{i}-\hat{j}+\hat{k})\left(\frac{\hat{i}-2 \hat{j}+\hat{k}}{\sqrt{6}}\right)= \pm \frac{2}{\sqrt{6}}= \pm \sqrt{\frac{4}{6}}= \pm \sqrt{\frac{2}{3}} \end{aligned}$

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 $Q$ 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} $

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

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

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

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

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

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.

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.

(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} $

Required equation of plane is

$62 x+29 y+19 z-105=0$

The required line is the intersection of planes

$2 x-y+z-3=0$ ..... (i)

and $3 x+y+z-5=0$ .... (ii)

Thus, the plane containing required line is

$(2 x-y+z-3)+\lambda(3 x+y+z-5)=0$

$\Rightarrow(2+3 \lambda) x+(\lambda-1) y+(\lambda+1) z+(-5 \lambda-3)=0$

As its distance from point $(2,1,-1)$ is $\frac{1}{\sqrt{6}}$

$\Rightarrow\left|\frac{(3 \lambda+2) 2+(\lambda-1) 1-(\lambda+1)+(-5 \lambda-3)}{\sqrt{11 \lambda^{2}+12 \lambda+6}}\right|=\frac{1}{\sqrt{6}}$

$\Rightarrow \frac{6 \lambda+4+\lambda-1-\lambda-1-5 \lambda-3}{\sqrt{11 \lambda^{2}+12 \lambda+6}}=\frac{1}{\sqrt{6}}$

$\Rightarrow 6(\lambda-1)^{2}=11 \lambda^{2}+12 \lambda+6$ (Squaring both side)

$\Rightarrow 6\left(\lambda^{2}+1-2 \lambda\right)=11 \lambda^{2}+12 \lambda+6$

$\Rightarrow \quad 5 \lambda^{2}+24 \lambda=0$

$\Rightarrow \quad \lambda(5 \lambda+24)=0$

$\Rightarrow \quad \lambda=0, \frac{-24}{5}$

Required planes are

$\begin{aligned} & \begin{array}{l} 2 x-y+z-3=0 \text { and } \\ 62 x+29 y+19 z-105=0 \\ \text { Or } \\ 3 x+y+z-5=0 \\ 62 x+29 y+19 z-105=0 \end{array} \\ \end{aligned}$