3D Geometry

Equation of the line joining points $ P(1, 2, -1) $ and $ Q(2, 3, 1) $ is

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

Let the equation of line $ L $ be as above. The point $ S $ is the foot of the perpendicular from $ R(4, -1, 5) $ to line $ L $.

The direction ratio of line $ L $ is $ (1, 1, 2) $.

We use the condition for perpendicularity, which says the dot product of direction ratios of the line and of the line joining $ R $ and $ S $ is zero.

So,

$ (\lambda - 3)(1) + (\lambda + 3)(1) + (2\lambda - 6)(2) = 0 $

Simplifying,

$ \lambda - 3 + \lambda + 3 + 4\lambda - 12 = 0 $

$ 6\lambda - 12 = 0 \Rightarrow \lambda = 2 $

Therefore, coordinates of point $ S $ are $ S(3, 4, 3) $.

Now, the point $ S $ divides the line segment $ PT $ internally in the ratio $ 1:2 $.

Using the section formula,

$ \frac{a + 2}{3} = 3, \quad \frac{b + 4}{3} = 4, \quad \frac{c - 2}{3} = 3 $

Hence,

$ a = 7, \ b = 8, \ c = 11 $

Therefore, $ T(7, 8, 11) $.

Now, the area of triangle $ PRT $ is given by

$ \text{Area} = \frac{1}{2} (PT)(RS) $

Substituting given values,

$ \text{Area} = \frac{1}{2} \times 6\sqrt{6} \times \sqrt{30} = 18\sqrt{5} $

Therefore, Option (D) is correct.

The equation of line $ RS $ is

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

Let point $ H $ lie on line $ RS $.

The direction ratios of $ PH $ are

$ \left( \frac{23}{5} - 1, -4 - 2, \frac{31}{5} + 1 \right) = \left( \frac{18}{5}, -6, \frac{36}{5} \right) $

The direction ratios of $ RT $ are $ (-3, -9, -6) $.

The dot product of $ \overrightarrow{PH} $ and $ \overrightarrow{RT} $ is

$ \overrightarrow{PH} \cdot \overrightarrow{RT} = -3\left(\frac{18}{5}\right) - 9(-6) - 6\left(\frac{36}{5}\right) $

$ = \frac{-54}{5} + 54 - \frac{216}{5} = 0 $

Since the dot product is zero, the two lines are perpendicular. Hence, point $ H $ is the orthocentre of triangle $ PRT $.

Therefore, Option (A) is correct.

The plane P contains the line

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

and is also perpendicular to the plane $x + 2y + 3z = 4.$

The direction ratios of line $L$ are $2, 3, 1.$
The normal vector of the given plane $x + 2y + 3z = 4$ is $1, 2, 3.$

Since plane $P$ is perpendicular to this plane, its normal vector $\vec{n}$ must be perpendicular to $(1, 2, 3)$ and also perpendicular to the line’s direction $(2, 3, 1).$

Thus, $\vec{n}$ is given by the cross product:

$\vec{n}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 3 \end{vmatrix}$

On simplifying,

$\vec{n}=7\hat{i}-5\hat{j}+\hat{k}$

The line passes through the point $(1, 3, -2)$, and since this point lies on the plane $P$, we use the point-normal form:

$7(x-1) - 5(y-3) + (z+2) = 0$

On simplifying,

$7x - 5y + z + (-7 + 15 + 2) = 0 \Rightarrow 7x - 5y + z = -10$

Hence, equation of plane $P$ is:

$7x - 5y + z = -10$

Option A is correct.

Plane $P_1$ is parallel to $P$. Hence it will have the same normal vector, so the general equation is:

$7x - 5y + z = d$

Since $P_1$ passes through the point $(4, 2, 2)$, substitute these coordinates:

$7(4) - 5(2) + 2 = d$

$28 - 10 + 2 = d \Rightarrow d = 20$

Therefore, equation of $P_1$ is

$7x - 5y + z = 20$

The distance between two parallel planes $A x + B y + C z + D_1 = 0$ and $A x + B y + C z + D_2 = 0$ is:

$\frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$

Here, $D_1 = -10$, $D_2 = -20$ (taking standard form). So,

$\text{Distance} = \frac{|-10 - 20|}{\sqrt{7^2 + (-5)^2 + 1^2}} = \frac{30}{\sqrt{75}}$

Option B is incorrect.

The formula for the perpendicular distance from point $(x_1, y_1, z_1)$ to plane $A x + B y + C z + D = 0$ is:

$\frac{|A x_1 + B y_1 + C z_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

For origin $(0, 0, 0)$:

$\text{Distance} = \frac{|0 + 0 + 0 - 10|}{\sqrt{75}} = \frac{10}{\sqrt{75}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

Option C is incorrect.

Acute angle between planes P and $2x + 2y + z = 3$

Let the normal vectors be:

$\vec{n}_1 = (7, -5, 1)$ and $\vec{n}_2 = (2, 2, 1)$.

The cosine of the angle between two planes is given by:

$\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$

Substitute the values:

$\cos\theta = \frac{|7(2) + (-5)(2) + 1(1)|}{\sqrt{75} \times 3} = \frac{1}{3\sqrt{3}}$

Hence,

$\theta = \cos^{-1}\left(\frac{1}{3\sqrt{3}}\right)$

Option D is correct.

To find the line of intersection $ L_1 $ of the two planes given by the equations $ 2x + 3y + z = 4 $ and $ x + 2y + z = 5 $, we use these equations to obtain the line in standard form:

$ \begin{aligned} & L_1: 2x + 3y + z = 4 \\ & x + 2y + z = 5 \end{aligned} $

Solving these equations gives us the line $ L_1 $ in standard form:

$ \frac{x+7}{1} = \frac{y-6}{-1} = \frac{z}{1} $

For the line $ L_2 $ that passes through the point $ P(2, -1, 3) $ and is parallel to $ L_1 $, the equation is:

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

The equation of the plane $ M $ is:

$ 2x + y - 2z = 6 $

To find the coordinates of the point $ Q $ where $ L_2 $ meets the plane $ M $, assume the coordinates of $ Q $ are $ (\lambda+2, -\lambda-1, \lambda+3) $. Since $ Q $ lies on the plane $ M $, substituting these into the equation of the plane gives:

$ 2(\lambda+2) + (-\lambda-1) - 2(\lambda+3) = 6 $

Solving, we find:

$ \lambda = -9 $

Thus, the coordinates of $ Q $ are $ (-7, 8, -6) $.

For the foot of the perpendicular $ R(x_1, y_1, z_1) $ from $ P $ to the plane $ M $, using the formula for the distance from a point to a plane:

$ \frac{x_1-2}{2} = \frac{y_1+1}{1} = \frac{z_1-3}{-2} = \frac{-(4-1-6-6)}{9} $

The solution provides the coordinates of $ R $ as $ (4, 0, 1) $.

Calculating the line segments:

$ PQ = 9\sqrt{3} $ units

$ QR = \sqrt{234} $ units

$ PR = 3 $ units

For the acute angle $ \theta $ between $ PQ $ and $ PR $:

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

$ \sin \theta = \sqrt{\frac{26}{27}} $

The area of triangle $ \triangle PQR $ is given by:

$ \text{Area} = \frac{1}{2} \times PQ \times PR \times \sin \theta = \frac{3}{2} \sqrt{234} \text{ square units} $

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

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

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

Now from (3)

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

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

$\begin{aligned} & \text { line }: \frac{x-1}{1}=\frac{y-3}{2}=\frac{z-2}{1} \\ & (x, y, z)=(\lambda+1,2 \lambda+3, \lambda+2) \\ & \text { Put in } L_1: x-y+3 z=6 \\ & (\lambda+1)-(2 \lambda+3)+3(\lambda+2)=6 \\ & 2 \lambda=2 \Rightarrow \lambda=1 \\ & Q=(2,5,3) \end{aligned}$

line: $\frac{x-2}{1}=\frac{y-5}{-1}=\frac{z-3}{3}$

$(\mathrm{x}, \mathrm{y}, \mathrm{z})=(\mathrm{t}+2,5-\mathrm{t}, 3 \mathrm{t}+3)$

Put in $L_2: 2 x-y+z=-4$

$\begin{aligned} & 2(t+4)-(5-t)+(3 t+3)=-4 \\ & 6 t=-6 \Rightarrow t=-1 \\ & R=(1,6,0) \end{aligned}$

$\begin{aligned} & \text { Perimeter }=\sqrt{6}+\sqrt{13}+\sqrt{11} \\ & \text { Centroid }=\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right) \end{aligned}$

$\begin{aligned} & \mathrm{S}=\left\{\mathrm{X}:(\mathrm{XP})^2-(\mathrm{XQ})^2=50\right\} \\ & \mathrm{T}=\left\{\mathrm{Y}:(\mathrm{YQ})^2-(\mathrm{YP})^2=50\right\} \end{aligned}$

for finding $S \equiv X(x, y, z)$ and for $T \equiv Y(x, y, z)$

$\begin{aligned} & \left((x-1)^2+(y-1)^2+(z-1)^2\right)-\left((x-4)^2+(y-2)^2+(z-7)^2\right)=50 \\ & \Rightarrow \quad S=\{(x, y, z): 6 x+8 z=105\} \\ & \quad \quad T=\{(x, y, z): 6 x+8 z=5\} \end{aligned}$

Since S and T both are plane;

(A) There exist a triangle in plane S whose area $=1$ (always)

(B) L & M lies on plane T, hence line segment joining L & M will lie on plane T.

(C) Distance between $\mathrm{S}$ & $\mathrm{T}$

$\mathrm{d}=\left|\frac{105-5}{10}\right|=10$

Hence for rectangle of perimeter 48 can exist.

There will be infinite such rectangle possible.

(D) For square

Hence Answers A, B, C, D are correct.

$\begin{aligned} & l_1: \vec{r}=\lambda(\hat{i}+\hat{j}+\hat{k}) \\\\ & l_2: \vec{r}=\hat{j}-\hat{k}+\mu(\hat{i}+\hat{k}) \\ & \end{aligned}$
For plane

$d(H)=$ Smallest possible distance between the points of $l_2$ and Plane.

$d\left(H_0\right)=$ Maximum value of $d(H)$

For $d\left(H_0\right)$



$l_2$ is Parallel to plane containing $l_1$

Equation of plane

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

$\because$ It contain $\mathrm{l}_1$

$ \therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=0 $ ..........(1)

For largest possible distance between plane (1) and $\mathrm{l}_2$ the line $\mathrm{l}_2$ must be parallel to plane (1)

$ \therefore \mathrm{a}+\mathrm{c}=0 $ .........(2)

By (1) and (2) $a=-c, b=0$

$\therefore$ Equation of plane $x-z=0$

(P) $ d\left(H_0\right)=P M=\left|\frac{0-(-1)}{\sqrt{1+1}}\right|=\frac{1}{\sqrt{2}}$

(Q) Distance from $(0,1,2)$

$ =\left|\frac{0-2}{\sqrt{2}}\right|=\sqrt{2} $

(R) Distance from origin $(0,0,0)$

$ =\left|\frac{0}{\sqrt{2}}\right|=0 $

(S) Point of Intersection,

$ x-z=0 $

and

$ \begin{aligned} & x=1, y=z \\\\ & \therefore x=1=z=y \end{aligned} $

$\therefore$ Point of intersection $(1,1,1)$ Distance from origin

$ =\sqrt{1^2+1^2+1^2}=\sqrt{3} $

$\mathrm{P}_1$ and $\mathrm{P}_2$ be two planes given by

$ \begin{aligned} & P_1: 10 x+15 y+12 z-60=0 \\\\ & P_2:-2 x+5 y+4 z-20=0 \end{aligned} $

Now finding line of intersection of both the planes,

Let $z=\lambda$, then

$ \begin{aligned} 10 x+15 y & =60-12 \lambda .........(1) \\\\ -2 x+5 y & =20-4 \lambda .........(2) \end{aligned} $

Now solving the eq. (1) and (2) we get,

$ \frac{x}{0}=\frac{y-4}{-4}=\frac{z}{5} \quad(\because \lambda=z) $

Now any skew line with the line of intersection of given plane can be edge of tertrahedron.

Now using above concept we will solve all options.

For option (A)

$ \frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{5} $

Point is $(1,1,5 \lambda+1)$

Now satisfying this point in given plane we have,

$ \begin{aligned} & 10 \times 1+15 \times 1+12 \times(5 \lambda+1)-60=0 \\\\ & \Rightarrow 60 \lambda=23 \\\\ & \Rightarrow \lambda=\frac{23}{60} \end{aligned} $

Now we can see line is intersecting the plane $\mathrm{P}_1$, at some point.

Now checking for plane $\left(\mathrm{P}_2\right)$

$ \begin{aligned} & -2 \times 1+5 \times 1 +4(5 \lambda+1)=20 \\\\ & \Rightarrow 20 \lambda =13 \\\\ & \Rightarrow \lambda =\frac{13}{20} \end{aligned} $

Also intersecting plane $\left(\mathrm{P}_2\right)$

Hence, it can be the edge of tetrahedron.

For option (B)

$ \frac{x-6}{-5}=\frac{y}{2}=\frac{z}{3} \text { point is }(-5 \lambda+6,2 \lambda, 3 \lambda) $

this point is sytisfying plane $P_1$

$ \begin{aligned} & 10(-5 \lambda+6)+15 \times 2 \lambda+12 \times 3 \lambda=60 \\\\ & \Rightarrow -50 \lambda+60+30 \lambda+36 \lambda=60 \\\\ & \Rightarrow 16 \lambda =0 \\\\ & \Rightarrow \lambda =0 \end{aligned} $

Now checking for plane $\mathrm{P}_2$

$ \begin{aligned} & -2(-5 \lambda+6)+5 \times 2 \lambda+4 \times 3 \lambda=20 \\\\ & \Rightarrow +10 \lambda-12+10 \lambda+12 \lambda=20 \\\\ & \Rightarrow 32 \lambda=32 \\\\ & \Rightarrow \lambda =1 \end{aligned} $

Hence, it can be the edge of tetrahedron.

For option (C)

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

point is $(-2 \lambda, 5 \lambda+4,4 \lambda)$

Similarly checking in plane $\mathrm{P}_1$ we get.

$ \begin{aligned} & \lambda \text { for } P_1=0 \\\\ & \lambda \text { for } P_2=0 \end{aligned} $

Hence, it can not be the edge of tetrahedron

For option (D),

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

point $(\lambda,-2 \lambda+4,3 \lambda)$ and for $\lambda=0$ point will be $(0,-4,0)$ which is lying on line of intersection and DR of plane $\mathrm{P}_2$ is $(-2,5,4)$ and DR of line is $(1,-2,3)$

Now line is lying completely on $\mathrm{P}_2$

Hence, it can be the edge of tetrahedron.

Given : equation of plane

$ \vec{r}=-(t+p) \hat{i}+t \hat{j}+(1+p) \hat{k} $

on rearranging we get,

$ \vec{r}=k+t+(-\hat{i}+\hat{j})+p(-\hat{i}+\hat{k}) $

So, equation of plane in standard form is given by

$ \begin{aligned} & {[\vec{r}-\hat{k}\,\,\,\,\,\,\,\,-\hat{i}+\hat{j}\,\,\,\,\,\,\,\,-\hat{i}+\hat{k}]=0} \\\\ & \Rightarrow\left[\begin{array}{rrr} x & y & z-1 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]=0 \\\\ & \Rightarrow x+y+z=1 .........(1) \end{aligned} $

Now given co-ordinate of $Q=(10,15,20)$

And Co-ordinates of $S=(\alpha, \beta, \gamma)$

Now using the image formula of point and plane we get,

$ \begin{aligned} & \frac{\alpha-10}{1}=\frac{\beta-15}{1}=\frac{\gamma-20}{1}=\frac{-2(10+15+20-1)}{3} \\\\ & \Rightarrow \alpha-10=\beta-15=\gamma-20=\frac{-88}{3} \\\\ & \Rightarrow \alpha=\frac{-58}{3}, \beta=\frac{-43}{3}, \gamma=\frac{-28}{3} \end{aligned} $

Now solving all options.

$ \begin{aligned} 3(\alpha+\beta) & =-101 \\\\ 3(\beta+\gamma) & =-71 \\\\ 3(\gamma+\alpha) & =-86 \\\\ 3(\alpha+\beta+\gamma) & =-129 \end{aligned} $

Since, the point A(3, 2, $-$1) is the mirror image of the point B(1, 0, $-$1) with respect to the plane $\alpha $x + $\beta $y + $\gamma $z = $\delta $, then

${\alpha \over {3 - 1}} = {\beta \over {2 - 0}} = {\gamma \over {( - 1) - ( - 1)}}$

$ \Rightarrow {\alpha \over 1} = {\beta \over 1} = {\gamma \over 0} \Rightarrow \gamma = 0$

it is given that $\alpha + \gamma = 1 \Rightarrow \alpha = 1$, so $\beta = 1$.

And, the mid-point of AB, M(2, 1, $-$1) lies on the given plane, so

$2\alpha + \beta - \gamma = \delta $

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

$ \therefore $ $\alpha + \beta = 2$, $\delta - \gamma = 0$, $\delta + \beta = 4$.

Equation of given straight lines

${L_1}:{{x - 1} \over 1} = {y \over { - 1}} = {{z - 1} \over 3}$

and ${L_2}:{{x - 1} \over { - 3}} = {y \over { - 1}} = {{z - 1} \over 1}$

having vector form respectively, are

and $\eqalign{ & r = (\widehat i + \widehat k) + \lambda (\widehat i - \widehat j + 3\widehat k) \cr & r = (\widehat i + \widehat k) + v( - 3\widehat i - \widehat j + \widehat k) \cr} $

$ \because $ $(\widehat i - \widehat j + 3\widehat k)\,.\,( - 3\widehat i - \widehat j + \widehat k)$

= $-$3 + 1 + 3 = 1 is positive,

$ \therefore $ Angle between supporting line vectors of lines L₁ and L₂ is acute, and point of intersection of given lines L₁ and L₂ is (1, 0, 1)

Now, vector along the acute angle bisector of vectors $(\widehat i - \widehat j + 3\widehat k)$ and $( - 3\widehat i - \widehat j + \widehat k)$ is $(\widehat i - \widehat j + 2\widehat k)$ or $(\widehat i + \widehat j - 2\widehat k)$

It is given that line

$L:{{x - \alpha } \over l} = {{y - 1} \over m} = {{z - \gamma } \over { - 2}}$ is the bisector of the acute angle between the lines L₁ and L₂, so

l = 1 and m = 1

and ${{I - \alpha } \over 1} = {{0 - 1} \over 1} = {{1 - \gamma } \over { - 2}}$

$ \Rightarrow \alpha = 2,\,\gamma = - 1$

$ \therefore $ $\alpha - \gamma = 3,\,l + m = 2$

Given lines,

${L_1}:r = \lambda \widehat i$, $\lambda $ $ \in $ R .......(i)

${L_2}:r = \widehat k + \mu \widehat j$, $\mu $ $ \in $ R .......(ii)

and ${L_3}:r = \widehat i + \widehat j + v\widehat k$, v $ \in $ R .......(iii)

Now, let the point P on L₁ = ($\lambda $, 0, 0)

the point Q on L₂ = (0, $\mu $, 1), and

the point R on L₃ = (1, 1, v)

For collinearity of points P, Q and R, there should be a non-zero scalar 'm', such that PQ = m PR

$ \Rightarrow ( - \lambda \widehat i + \mu \widehat j + \widehat k) = m[(1 - \lambda )\widehat i + \widehat j + v\widehat k]$

$ \Rightarrow {\lambda \over {\lambda - 1}} = {\mu \over 1} = {1 \over v}$

$ \Rightarrow v = {1 \over \mu }$ and $\lambda = {\mu \over {\mu - 1}}$

where, $\mu \ne 0$ and $\mu \ne 1$

$ \Rightarrow $ $Q \ne \widehat k$ and $Q \ne \widehat k + \widehat j$

Hence, Q can not have coordinator (0, 0, 1 ) and (0, 1, 1)

Hence, options (c) and (d) are correct.

Given lines

${L_1}:r = \widehat i + \lambda ( - \widehat i + 2\widehat j + 2\widehat k),\,\lambda \in R$ and

${L_2}:r = \mu (2\widehat i - \widehat j + 2\widehat k),\,\mu \in R$

and since line L₃ is perpendicular to both lines L₁ and L₂.

Then a vector along L₃ will be,

$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1} & 2 & 2 \cr 2 & { - 1} & 2 \cr } } \right| = \widehat i(4 + 2) - \widehat j( - 2 - 4) + \widehat k(1 - 4)$

$ = 6\widehat i + 6\widehat j - 3\widehat k = 3(2\widehat i + 2\widehat j - \widehat k)$ ....(i)

Now, let a general point on line L₁.

$P(1 - \lambda ,\,2\lambda ,\,2\lambda )$ and on line L₂ as $Q(2\mu - \mu ,\,2\mu )$ and let P and Q are point of intersection of lines L₁, L₃ and L₂, L₃, so direction ratio's of L₃

$(2\mu + \lambda - 1,\, - \mu - 2\lambda ,\,\,2\mu - 2\lambda )$ ....(ii)

Now, ${{2\mu + \lambda - 1} \over 2} = {{\, - \mu - 2\lambda } \over 2}\, = {{\,2\mu - 2\lambda } \over 1}$

[from Eqs. (i) and (ii)]

$ \Rightarrow \lambda = {1 \over 9}$ and $\mu = {2 \over 3}$

So, $P\left( {{8 \over 9},{2 \over 9},{2 \over 9}} \right)$ and $Q\left( {{4 \over 9}, - {2 \over 9},{4 \over 9}} \right)$

Now, we can take equation of line L₃ as $r = a + t(2\widehat i + 2\widehat j - \widehat k)$, where a is position vector of any point on line L₃ and possible vector of a are

$\left( {{8 \over 9}\widehat i + {2 \over 9}\widehat j + {2 \over 9}\widehat k} \right)$ or $\left( {{4 \over 9}\widehat i - {2 \over 9}\widehat j + {4 \over 9}\widehat k} \right)$ or $\left( {{2 \over 3}\widehat i + {1 \over 3}\widehat k} \right)$

Hence, options (a), (b) and (c) are correct.

We have,

and $\eqalign{ & {P_1}:2x + y - z = 3 \cr & {P_2}:x + 2y + z = 2 \cr} $

Here, $\overrightarrow {{n_1}} = 2\widehat i + \widehat j - \widehat k$

and $\overrightarrow {{n_2}} = \widehat i + 2\widehat j + \widehat k$

(a) Direction ratio of the line of intersection of P₁ and P₂ is $\theta $ $\overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $

i.e. $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & { - 1} \cr 1 & 2 & 1 \cr } } \right|$

$ = (1 + 2)\widehat i - (2 + 1)\widehat j + (4 - 1)\widehat k$

$ = 3(\widehat i - \widehat j + \widehat k)$

Hence, statement a is false.

(b) We have,

${{3x - 4} \over 9} = {{1 - 3y} \over 9} = {z \over 3}$

$ \Rightarrow {{x - {4 \over 3}} \over 3} = {{\left( {y - {1 \over 3}} \right)} \over { - 3}} = {z \over 3}$

This line is parallel to the line of intersection of P₁ and P₂.

Hence, statement (b) is false.

(c) Let acute angle between P₁ and P₂ be $\theta $.

We know that,

$\cos \theta = {{\overrightarrow {{n_1}} .\overrightarrow {{n_2}} } \over {|\overrightarrow {{n_1}} ||\overrightarrow {{n_2}} |}}$

$ = {{(2\widehat i + \widehat j - \widehat k).(\widehat i + 2\widehat j + \widehat k)} \over {|2\widehat i + \widehat j - \widehat k||\widehat i + 2\widehat j + \widehat k|}}$

$ = {{2 + 2 - 1} \over {\sqrt 6 \times \sqrt 6 }} = {1 \over 2}$

$\theta = 60^\circ $

Hence, statement (c) is true.

(d) Equation of plane passing through the point (4, 2, $-$2) and perpendicular to the line of intersection of P₁ and P₂ is

$\eqalign{ & 3(x - 4) - 3(y - 2) + 3(z + 2) = 0 \cr & \Rightarrow 3x - 3y + 3z - 12 + 6 + 6 = 0 \cr} $

$ \Rightarrow 3x - 3y + 3z - 12 + 6 + 6 = 0$

$ \Rightarrow x - y + z = 0$

Now, distance of the point (2, 1, 1) from the plane x $-$ y + z = 0 is

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

Hence, statement (d) is true.

Let the equation of plane be ax + by + cz = 1. Then

a + b + c = 1

2a + b $-$ 2c = 0

3a $-$ 6b $-$ 2c = 0

$ \Rightarrow $ a = 7b

c = ${{15b} \over 2}$

b = ${{2} \over 31}$, a = ${{14} \over 31}$, c = ${{15} \over 31}$

$ \therefore $ 14x + 2y + 15z = 31

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.

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 $\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

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 $

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

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

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.

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

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} $

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} $

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

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