3D Geometry

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

$(3 \lambda+2,-2 \lambda-1,-\lambda-3)$ lies on plane $p x-q y+z=5$

$p(3 \lambda+2)-q(-2 \lambda-1)+(-\lambda-3)=5$

$\lambda(3 p+2 q-1)+(2 p+q-8)=0$

$3 p+2 q-1=0\} p=15$

$2 p+q-8=0\} q=-22$

Equation of plane $15 x+22 y+z-5=0$

Shortest distance from origin $=\frac{|0+0+0-5|}{\sqrt{15^{2}+22^{2}+1}}$

$=\frac{5}{\sqrt{710}}$

$=\sqrt{\frac{5}{142}}$

$P(1,2,1)$ image in plane $x+2 y+2 z=16$

$ \begin{aligned} & \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=\frac{-2(1+2 \times 2+2 \times 1-16)}{1^{2}+2^{2}+2^{2}} \\\\ & \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=2 \\\\ & Q(3,6,5) \\\\ & \vec{r}=-\hat{k}+\lambda(\hat{i}+\hat{j}+2 \hat{k}) \end{aligned} $


$A Q=3 \hat{i}+6 \hat{j}+6 \hat{k}$

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

Equation of plane $\equiv 2(x-0)+0(y-0)-1(z+1)$ $=0$

$2 x-z=1$

Point lying on plane from the option is $(1,2,1)$ i.e., option (B)

We know mirror image of point (x₁, y₁, z₁) in the plane ax + by + cz = d

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

Here given point (2, 4, 7) and plane $3x - y + 4z = 2$ then mirror image is

${{x - 2} \over 3} = {{y - 4} \over { - 1}} = {{z - 7} \over 4} = {{ - 2(6 - 4 + 28 - 2)} \over {9 + 1 + 16}}$

$ \Rightarrow {{x - 2} \over 3} = {{y - 4} \over { - 1}} = {{z - 7} \over 4} = - {{28} \over {13}}$

$\therefore$ $x = - {{58} \over {13}} = a$

$y = {{80} \over {13}} = b$

$z = - {{21} \over {13}} = c$

$\therefore$ $2a + b + 2c$

$ = 2\left( { - {{58} \over {13}}} \right) + {{80} \over {13}} + 2\left( { - {{21} \over {13}}} \right)$

$ = {{ - 116 + 80 - 42} \over {13}} = {{ - 78} \over {13}} = - 6$

Equation of pane through point (2, 3, $-$5) and perpendicular to planes 2x + y $-$ 5z = 10 and 3x + 5y $-$ 7z = 12 is

$\left| {\matrix{ {x - 2} & {y - 3} & {z + 5} \cr 2 & 1 & { - 5} \cr 3 & 5 & { - 7} \cr } } \right| = 0$

$\therefore$ Equation of plane is $(x - 2)( - 7 + 25) - (y - 3)$

$( - 14 + 15) + (z + 5)\,.\,7 = 0$

$\therefore$ $18x - y + 7z + 2 = 0$

$ \Rightarrow 18x - y + 7z = - 2$

$\therefore$ $ - 18x + y - 7z = 2$

On comparing with $ax + by + cz = d$ where d > 0 is a = $-$ 18, b = 1, c = $-$ 7, d = 2

$\therefore$ $a + 7b + c + 20d = 22$

Line L is x = y = z

$\overrightarrow {PQ} .\,(\widehat i + \widehat j + \widehat k) = 0$

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

$ \Rightarrow \alpha = {2 \over 3}$ so, $T = \left( {{2 \over 3},{2 \over 3},{2 \over 3}} \right)$

$PT = \sqrt {{{38} \over 3}} $

$ \Rightarrow QT = {4 \over {\sqrt 3 }}$

So, Area $ = \left( {{1 \over 2} \times {4 \over {\sqrt 3 }} \times {{\sqrt {38} } \over {\sqrt 3 }}} \right).\,2$

$ = {{4\sqrt {38} } \over 3}$ sq. units

Family of Plane's equation can be given by

$(5 + 8\lambda )x + (8 - 7\lambda )y + (13 + \lambda )z - (29 + 20\lambda ) = 0$

P₁ passes through (2, 1, 3)

$ \Rightarrow (10 + 16\lambda ) + (8 - 7\lambda ) + (39 + 3\lambda ) - (29 + 20\lambda ) = 0$

$ \Rightarrow - 8\lambda + 28 = 0 \Rightarrow \lambda = {7 \over 2}$

d.r, s of normal to P₁

$\left\langle {33,{{ - 33} \over 2},{{33} \over 2}} \right\rangle $ or $\left\langle {1, - {1 \over 2},{1 \over 2}} \right\rangle $

P₂ passes through (0, 1, 2)

$ \Rightarrow 8 - 7\lambda + 26 + 2\lambda - (29 + 20\lambda ) = 0$

$ \Rightarrow 5 - 25\lambda = 0$

$ \Rightarrow \lambda = {1 \over 5}$

d.r, s of normal to P₂

$\left\langle {{{33} \over 5},{{33} \over 5},{{66} \over 5}} \right\rangle $ or $\left\langle {1,1,2} \right\rangle $

Angle between normals

$ = {{\left( {\widehat i - {1 \over 2}\widehat j + {1 \over 2}\widehat k} \right).\,\left( {\widehat i + \widehat j + 2\widehat k} \right)} \over {{{\sqrt 3 } \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt 6 }}$

$\cos \theta = {{1 - {1 \over 2} + 1} \over 3} = {1 \over 2}$

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

${P_1}:x + 3y - z = 6$

${P_2}: - 6x + 5y - z = 7$

Family of planes passing through line of intersection of P₁ and P₂ is given by $x(1 - 6\lambda ) + y(3 + 5\lambda ) + z( - 1 - \lambda ) - (6 + 7\lambda ) = 0$

It passes through $\left( {2,3,{1 \over 2}} \right)$

So, $2(1 - 6\lambda ) + 3(3 + 5\lambda ) + {1 \over 2}( - 1 - \lambda ) - (6 + 7\lambda ) = 0$

$ \Rightarrow 2 - 12\lambda + 9 + 15\lambda - {1 \over 2} - {\lambda \over 2} - 6 - 7\lambda = 0$

$ \Rightarrow {9 \over 2} - {{9\lambda } \over 2} = 0 \Rightarrow \lambda = 1$

Required plane is

$ - 5x + 8y - 2z - 13 = 0$

Or $\overrightarrow r .\,( - 5\widehat i + 8\widehat j - 2\widehat k) = 13$

${{|13\overrightarrow a {|^2}} \over {|d{|^2}}} = {{{{13}^2}} \over {{{(13)}^2}}}.\,|\overrightarrow a {|^2} = 93$

$L:{{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3} = t$

Let P = (4t $-$ 2, 2t + 1, 3t $-$ 1)

$\because$ P is the foot of perpendicular of (1, 2, 4)

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

$ \Rightarrow 29t = 29 \Rightarrow t = 1$

$\therefore$ P = (2, 3, 2)

Now, distance of P from the plane

$3x + 4y + 12z + 23 = 0$, is

$\left| {{{6 + 12 + 24 + 23} \over {\sqrt {9 + 16 + 144} }}} \right| = {{65} \over {13}} = 5$

${L_1}:{{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$

${L_2}:{{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$

Now, $\overrightarrow p \times \overrightarrow q = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 3 & { - 1} \cr 2 & 1 & 3 \cr } } \right| = 10\widehat i - 8\widehat j - 4\widehat k$

and ${\overrightarrow a _2} - {\overrightarrow a _1} = 6\widehat i - 4\widehat j - 4\widehat k$

$\therefore$ $S.D. = \left| {{{60 + 32 + 16} \over {\sqrt {100 + 64 + 16} }}} \right| = {{108} \over {\sqrt {180} }} = {{18} \over {\sqrt 5 }}$

Given that the direction cosines satisfy $l + m - n = 0$, we find that $n = l + m$.

The other equation is $3l^2 + m^2 + cnl = 0$, and substituting $n = l + m$ gives $3l^2 + m^2 + cl(l + m) = 0$.

This simplifies to $(3 + c)l^2 + clm + m^2 = 0$.

As the lines are parallel, they share the same direction ratios, so we can express $l$ in terms of $m$, say $l = km$. Substituting this into our equation gives $(3 + c)(km)^2 + ckm^2 + m^2 = 0$.

This simplifies to $m^2[k^2(3 + c) + kc + 1] = 0$.

Since $m \neq 0$, we must have $k^2(3 + c) + kc + 1 = 0$. Here, we consider the ratio $k = \frac{l}{m}$ to be constant, since the lines are parallel. The equation then becomes a quadratic equation in $k$.

As the lines are parallel, the discriminant of the quadratic equation must be equal to zero for the equation to have equal roots. Hence, the discriminant $D = (c^2 - 4(3 + c)) = 0$.

Solving this quadratic equation gives $c^2 - 4c - 12 = 0$.

Factoring this equation gives $(c - 6)(c + 2) = 0$.

Solving for $c$ gives $c = 6, -2$.

However, we are looking for the positive value of $c$, so $c = 6$.

Therefore, the correct answer is 6

${P_1}:2x + y - 52 = 0$, ${P_2}:3x - y + 4z - 7 = 0$

Family of planes P₁ and P₂

$P:{P_1} + \lambda {P_2}$

$\therefore$ $P:(2 + 3\lambda )x + (1 - \lambda )y + ( - 5 + 4\lambda )z - 7\lambda = 0$

$\because$ $P \bot {P_1}$

$\therefore$ $4 + 6\lambda + 1 - \lambda + 25 - 20\lambda = 0$

$\lambda = 2$

$\therefore$ $P:8x - y + 32 - 14 = 0$

It passes through the point (1, 0, 2)

$\because$ Both lines are coplanar, so

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

$ \Rightarrow \alpha = {5 \over 3}$

Equation of plane containing both lines

$\left| {\matrix{ {x - 1} & {y + 1} & {z - 1} \cr 0 & 3 & { - 1} \cr 2 & 0 & { - 3} \cr } } \right| = 0$

$ \Rightarrow 9x + 2y + 6z = 13$

So, distance of $\left( {{5 \over 3},0,0} \right)$ from this plane

$ = {2 \over {\sqrt {81 + 4 + 36} }} = {2 \over {11}}$

Let $\overrightarrow v = {\lambda _1}\overrightarrow a + {\lambda _2}\overrightarrow b $, where ${\lambda _1},\,{\lambda _2} \in R$.

$ = ({\lambda _1} + 2{\lambda _2})\widehat i + ({\lambda _1} - 3{\lambda _2})\widehat j + (2{\lambda _1} + {\lambda _2})\widehat k$

$\because$ Projection of $\overrightarrow v $ on $\overrightarrow c $ is ${2 \over {\sqrt 3 }}$

$\therefore$ ${{{\lambda _1} + 2{\lambda _2} - {\lambda _1} + 3{\lambda _2} + 2{\lambda _1} + {\lambda _2}} \over {\sqrt 3 }} = {2 \over {\sqrt 3 }}$

$\therefore$ ${\lambda _1} + 3{\lambda _2} = 1$ ..... (i)

and $\overrightarrow v \,.\,\widehat j = 7 \Rightarrow {\lambda _1} - 3{\lambda _2} = 7$ ... (ii)

from equation (i) and (ii)

${\lambda _1} = 4$, ${\lambda _2} = - 1$

$\therefore$ $\overrightarrow v = 2\widehat i + 7\widehat j + 7\widehat k$

$\therefore$ $\overrightarrow v \,.\,(\widehat i + \widehat k) = 2 + 7$

$ = 9$

$\because$ L₁ and L₂ are perpendicular, so

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

$ \Rightarrow \alpha = 3$

Now angle between l₂ and l₃,

$\cos \theta = {{1( - 3) + {\alpha \over 2}( - 2) + 2(4)} \over {\sqrt {1 + {{{\alpha ^2}} \over 4} + } 4\sqrt {9 + 4 + 16} }}$

$ \Rightarrow \cos \theta = {2 \over {{{29} \over 2}}} \Rightarrow \theta = {\cos ^{ - 1}}\left( {{4 \over {29}}} \right) = {\sec ^{ - 1}}\left( {{{29} \over 4}} \right)$

Consider the equation of plane,

$P:(2x + 3y + z + 20) + \lambda (x - 3y + 5z - 8) = 0$

$P:(2 + \lambda )x + 3(3 - 3\lambda )y + 1(1 + 5\lambda )z + (20 - 8\lambda ) = 0$

$\because$ Plane P is perpendicular to $2x + 3y + z + 20 = 0$

So, $4 + 2\lambda + 9 - 9\lambda + 1 + 5\lambda = 0$

$ \Rightarrow \lambda = 7$

$P:9x - 18y + 36z - 36 = 0$

or $P:x - 2y + 4z = 4$

If image of $\left( {2, - {1 \over 2},2} \right)$ in plane P is (a, b, c) then

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

and $\left( {{{a + 2} \over 2}} \right) - 2\left( {{{b - {1 \over 2}} \over 2}} \right) + 4\left( {{{c + 2} \over 2}} \right) = 4$

Clearly $a = {4 \over 3}$, $b = {5 \over 6}$ and $c = - {2 \over 3}$

So, $a:b:c = 8:5: - 4$

Let the equation of required plane

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

$\because$ (2, 1, $-$2) lies on it so, $2 + \lambda ( - 2) = 0$

$ \Rightarrow \lambda = 1$

Hence, $\pi :3x + 2y - 8 = 0$

$\because$ ${\pi _x} = - 9$, ${\pi _y} = 5$, ${\pi _{x + y}} = 4$

${\pi _{x - y}} = - 22$ and ${\pi _{y - x}} = 6$

Clearly X and Y are on opposite sides of plane $\pi$

As L is parallel to PQ d.r.s of S is <1, 1, 1>

$\therefore$ $L \equiv {{x - 1} \over 1} = {{y + 1} \over 1} = {{z + 1} \over 1}$

Point of intersection of L and S be $\lambda$

$ \Rightarrow (\lambda + 1) + (\lambda - 1) + (\lambda - 1) = S$

$ \Rightarrow \lambda = 2$

$\therefore$ $R \equiv (3,1,1)$

Let $Q(\alpha ,\beta ,\gamma )$

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

$ \Rightarrow \alpha = 3,\,\beta = 2,\,\gamma = 3$

$ \Rightarrow Q \equiv (3,2,3)$

${(QR)^2} = {0^2} + {(1)^2} + {(2)^2} = 5$

Let ${\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat k$

${\overrightarrow a _2} = 2\widehat i + 4\widehat j + 5\widehat k$

$\overrightarrow p = 2\widehat i + 3\widehat j + \lambda \widehat k,\,\overrightarrow q = \widehat i + 4\widehat j + 5\widehat k$

$\therefore$ $\overrightarrow p \times \overrightarrow q = (15 - 4\lambda )\widehat i - (10 - \lambda )\widehat j + 5\widehat k$

${\overrightarrow a _2} - {\overrightarrow a _1} = \widehat i + 2\widehat j + 2\widehat k$

$\therefore$ Shortest distance

$ = \left| {{{(15 - 4\lambda ) - 2(10 - \lambda ) + 10} \over {\sqrt {{{(15 - 4\lambda )}^2} + {{(10 - \lambda )}^2} + 25} }}} \right| = {1 \over {\sqrt 3 }}$

$ \Rightarrow 3{(5 - 2\lambda )^2} = {(15 - 4\lambda )^2} + {(10 - \lambda )^2} + 25$

$ \Rightarrow 5{\lambda ^2} - 80\lambda + 275 = 0$

$\therefore$ Sum of values of $\lambda = {{80} \over 5} = 16$

Let P(x, y, z) be any point on plane P₁

Then ${(x + 4)^2} + {(y - 2)^2} + {(z - 1)^2} = {(x - 2)^2} + {(y + 2)^2} + {(z - 3)^2}$

$ \Rightarrow 12x - 8y + 4z + 4 = 0$

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

And ${P_2}:2x + y + 3z = 0$

$\therefore$ angle between P₁ and P₂

$\cos \theta =\left| {{{6 - 2 + 3} \over {14}}} \right| \Rightarrow \theta = {\pi \over 3}$

$\mathrm{P}_1$ 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} $

${P_1}:x - 2y - 2z + 1 = 0$

${P_2}:2x - 3y - 6z + 1 = 0$

$\left| {{{x - 2y - 2z + 1} \over {\sqrt {1 + 4 + 4} }}} \right| = \left| {{{2x - 3y - 6z + 1} \over {\sqrt {{2^2} + {3^2} + {6^2}} }}} \right|$

${{x - 2y - 2z + 1} \over 3} = \pm {{2x - 3y - 6z + 1} \over 7}$

Since ${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 20 > 0$

$\therefore$ Negative sign will give acute bisector

$7x - 14y - 14z + 7 = - [6x - 9y - 18z + 3]$

$ \Rightarrow 13x - 23y - 32z + 10 = 0$

$\left( { - 2,0, - {1 \over 2}} \right)$ satisfy it $\therefore$ Ans. (b)

$3y - 2z - 1 = 0 = 3x - z + 4$

$3y - 2z - 1 = 0$

D.R's $\Rightarrow$ (0, 3, $-$2)

$3x - z + 4$ = 0

D.R's $\Rightarrow$ (3, $-$1, 0)

Let DR's of given line are a, b, c

Now, 3b $-$ 2c = 0 & 3a $-$ c = 0

$\therefore$ 6a = 3b = 2c

a : b : c = 3 : 6 : 9

Any point on line

3K $-$ 1, 6K + 1, 9K + 1

Now, 3(3K $-$ 1) + 6(6K + 1)1 + 9(9K + 1) = 0

$\Rightarrow$ K = ${1 \over 3}$

Point on line $\Rightarrow$ (0, 3, 4)

Given point (2, $-$1, 6)

$\Rightarrow$ Distance = $\sqrt {4 + 16 + 4} = 2\sqrt 6 $

Option (c)

P₁ : 2x + 3y + 2z = 0

$\Rightarrow$ ${\overrightarrow n _1} = 2\widehat i + 3\widehat j + 2\widehat k$

P₂ : x $-$ 2y + z = 0

$\Rightarrow$ ${\overrightarrow n _2} = \widehat i - 2\widehat j + \widehat k$

Direction vector of line L which is line of intersection of P₁ & P₂

$\overrightarrow r = {\overrightarrow n _1} \times {\overrightarrow n _2} = 7\widehat i - 7\widehat k$

DR's of L are (1, 0, $-$1)

$\Rightarrow$ Equation of L : ${x \over 1} = {y \over 0} = {z \over { - 1}} = \lambda $



DR's of $\overrightarrow {PQ} $ = ($\lambda$ + 1, $-$2, 2 $-$ $\lambda$)

$\because$ $\overrightarrow {PQ} \bot \overrightarrow r $

$ \Rightarrow (\lambda + 1)(1) + ( - 2)(0) + (2 - \lambda )( - 1) = 0$

$ \Rightarrow \lambda = {1 \over 2} \Rightarrow Q\left( {{1 \over 2},0,{{ - 1} \over 2}} \right)$

$ \Rightarrow PQ = {{\sqrt {34} } \over 2}$

3x $-$ 2y + 4z $-$ 7 + $\lambda$(x + 5y $-$ 2z + 9) = 0

(3 + $\lambda$)x + (5$\lambda$ $-$ 2)y + (4 $-$ 2$\lambda$)z + 9$\lambda$ $-$ 7 = 0

passing through (1, 4, $-$3)

$\Rightarrow$ 3 + $\lambda$ + 20$\lambda$ $-$ 8 $-$ 12 + 6$\lambda$ + 9$\lambda$ $-$ 7 = 0

$\Rightarrow$ $\lambda$ = ${2 \over 3}$

$\Rightarrow$ equation of plane is

$-$11x $-$ 4y $-$ 8z + 3 = 0

$\Rightarrow$ $\alpha$ + $\beta$ + $\gamma$ = $-$23

n = 2 (l + m)

lm + n(l + m) = 0

lm + 2(l + m)² = 0

2l² + 2m² + 5ml = 0

$2{\left( {{l \over m}} \right)^2} + 2 + 5\left( {{l \over m}} \right) = 0$

2t² + 5t + 2 = 0

(t + 2)(2t + 1) = 0

$ \Rightarrow t = - 2; - {1 \over 2}$

(i) ${l \over m} = - 2$

${n \over m} = - 2$

($-$2m, m, $-$2m)

($-$2, 1, $-$2)

(ii) ${l \over m} = - {1 \over 2}$

n = $-$2l

(l, $-$2l, $-$2l)

(1, $-$2, $-$2)

$\cos \theta = {{ - 2 - 2 + 4} \over {\sqrt 9 \sqrt 9 }} = 0 \Rightarrow 0 = {\pi \over 2}$

Equation of planes are

$\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) - 1 = 0 \Rightarrow x + y + z - 1 = 0$

and $\overrightarrow r .\left( {2\widehat i + 3\widehat j - \widehat k} \right) + 4 = 0 \Rightarrow 2x + 3y - z + 4 = 0$

equation of planes through line of intersection of these planes is :-

$(x + y + z - 1) + \lambda (2x + 3y - z + 4) = 0$

$ \Rightarrow (1 + 2\lambda )x + (1 + 3\lambda )y + (1 - \lambda )z - 1 + 4\lambda = 0$

But this plane is parallel to x-axis whose direction are (1, 0, 0)

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

$\lambda = - {1 \over 2}$

$\therefore$ Required plane is

$0x + \left( {1 - {3 \over 2}} \right)y + \left( {1 + {1 \over 2}} \right)z - 1 + 4\left( {{{ - 1} \over 2}} \right) = 0$

$ \Rightarrow {{ - y} \over 2} + {3 \over 2}z - 3 = 0$

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

$ \Rightarrow \overrightarrow r .\left( {\widehat j - 3\widehat k} \right) + 6 = 0$ Ans.

$(1 + 2\lambda ) + 2 - 3\lambda + 3 - 6\lambda = 5$

$ \Rightarrow 6 - 7\lambda = 5 \Rightarrow \lambda = {1 \over 7}$

so, $P = \left( {{9 \over 7}, - {{11} \over 7},{{15} \over 7}} \right)$

$AP = \sqrt {{{\left( {1 - {9 \over 7}} \right)}^2} + {{\left( { - 2 + {{11} \over 7}} \right)}^2} + {{\left( {3 - {{15} \over 7}} \right)}^2}} $

$AP = \sqrt {\left( {{4 \over {49}}} \right) + {9 \over {49}} + {{36} \over {49}}} = 1$

Required equation of plane

${P_1} + \lambda {P_2} = 0$

$(x - y - z - 1) + \lambda (2x + y - 3z + 4) = 0$

Given that its dist. From origin is ${2 \over {\sqrt {21} }}$

Thus, ${{|4\lambda - 1|} \over {\sqrt {{{(2\lambda + 1)}^2} + {{(\lambda - 1)}^2} + {{( - 3\lambda - 1)}^2}} }} = {{\sqrt 2 } \over {\sqrt {21} }}$

$ \Rightarrow 21{(4\lambda - 1)^2} = 2(14{\lambda ^2} + 8\lambda + 3)$

$ \Rightarrow 336{\lambda ^2} - 168\lambda + 21 = 28{\lambda ^2} + 16\lambda + 6$

$ \Rightarrow 308{\lambda ^2} - 184\lambda + 15 = 0$

$ \Rightarrow 308{\lambda ^2} - 154\lambda - 30\lambda + 15 = 0$

$ \Rightarrow (2\lambda - 1)(154\lambda - 15) = 0$

$ \Rightarrow \lambda = {1 \over 2}$ or ${{15} \over {154}}$

for $\lambda = {1 \over 2}$ reqd. plane is $4x - y - 5z + 2 = 0$

$(x + y + 4z - 16) + \lambda ( - x + y + z - 6) = 0$

Passes through (1, 2, 3)

$ - 1 + \lambda ( - 2) \Rightarrow \lambda = - {1 \over 2}$

$2(x + y + 4z - 16) - ( - x + y + z - 6) = 0$

$3x + y + 7z - 26 = 0$

Equation of plane P can be assumed as

P : x + 2y + 3z + 1 + $\lambda$ (x $-$ y $-$ z $-$ 6) = 0

$\Rightarrow$ P : (1 + $\lambda$)x + (2 $-$ $\lambda$)y + (3 $-$ $\lambda$)z + 1 $-$ 6$\lambda$ = 0

$ \Rightarrow {\overrightarrow n _1} = (1 + \lambda )\widehat i + (2 - \lambda )\widehat j + (3 - \lambda )\widehat k$

$\therefore$ ${\overrightarrow n _1}\,.\,{\overrightarrow n _2} = 0$

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

$\Rightarrow$ 2 + 2$\lambda$ $-$ 2 + $\lambda$ $-$ 3 + $\lambda$ = 0 $\Rightarrow$ $\lambda$ = ${3 \over 4}$

$\Rightarrow$ $P:{{7x} \over 4} + {5 \over 4}y + {{9z} \over 4} - {{14} \over 4} = 0$

$\Rightarrow$ 7x + 5y + 9z = 14

(0, 1, 1) lies on P.

First line is ($\phi$ + $\alpha$, 2$\phi$ + 1, 3$\phi$ + 1)

and second line is (q$\beta$ + 4, 3q + 6, 3q + 7)

For intersection

$\phi$ + $\alpha$ = q$\beta$ + 4 ...... (i)

2$\phi$ + 1 = 3q + 6 .... (ii)

3$\phi$ + 1 = 3q + 7 ...... (iii)

for (ii) & (iii) $\phi$ = 1, q = $-$1

So, from (i) $\alpha$ + $\beta$ = 3

Now, point of intersection is ($\alpha$ + 1, 3, 4)

It lies on the plane.

Hence, $\alpha$ = 5 & $\beta$ = $-$2

Normal of required plane $\left( {2\widehat i + \widehat j - \widehat k} \right) \times \left( {\widehat i - \widehat j - \widehat k} \right)$

$ = - 2\widehat i + \widehat j - 3\widehat k$

Equation of plane

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

$ - 2x + y - 3z - 8 = 0$

$2x - y + 3z + 8 = 0$

$a + b + c = 4$

$\overrightarrow {PN} .(\widehat i - \widehat k) = 0$

$\Rightarrow$ N(1, 0, $-$1)

Now,


$\overrightarrow {PQ} .(\widehat i + \widehat j + 2\widehat k) = 0$

$\Rightarrow$ $\mu$ = $-$ 1

$\Rightarrow$ Q ($-$1, 0, 1)

$\overrightarrow {PN} $ = 2$\widehat j$ and $\overrightarrow {PQ} $ = $2\widehat i + 2\widehat j - 2\widehat k$

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