3D Geometry

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} $

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

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.

$\mathrm{P}_1$ 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 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 $\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

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, 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.

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

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} $