3D Geometry

$l = {1 \over 2},m = {1 \over {\sqrt 2 }},n = \cos \theta $

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

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

$\theta$ is acute $\therefore$ $n = {1 \over 2}$

$\therefore$ $\overrightarrow v = k\left( {{1 \over 2}\widehat i + {1 \over {\sqrt 2 }}\widehat j + {1 \over 2}\widehat k} \right),k \in R$

$\overrightarrow v .\,(\overrightarrow a - \overrightarrow b ) = 0$

$(\sqrt 2 - a){1 \over 2} + ( - 1 - b){1 \over {\sqrt 2 }} + (1 - c){1 \over 2} = 0$

$ \Rightarrow {a \over 2} + {b \over {\sqrt 2 }} + {c \over 2} = {1 \over 2}$

$ \Rightarrow a + \sqrt 2 b + c = 1$

Let $P \equiv ( - 1,k,0),Q \equiv (2,k, - 1)$ & $R(1,1,2)$

$\overrightarrow P R = 2\widehat i + (1 - k)\widehat j + 2\widehat k$

& $\overrightarrow Q R = - \widehat i + (1 - k)\widehat j + 3\widehat k$

$\therefore$ Normal to plane will be

$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & {(1 - k)} & 2 \cr { - 1} & {(1 - k)} & 3 \cr } } \right| = \widehat i(1 - k) - \widehat j(8) + 3\widehat k(1 - k)$

If line is parallel to this we have

$1(1 - k) + 1( - 8) + ( - 3)(1 - k) = 0$

$ \Rightarrow 2(1 - k) = - 8$

$ \Rightarrow 1 - k = - 4 \Rightarrow k = 5$

$\therefore$ ${{{k^2} + 1} \over {(k - 1)(k - 2)}} = {{26} \over {4.3}} = {{13} \over 6}$

Equation of ${l_1} = {{x - 2} \over 2} = {{y - 6} \over 1} = {{z - 2} \over { - 2}}$

Shortest distance with ${{x + 1} \over 2} = {{y + 4} \over { - 3}} = {z \over 2}$ is

S.d $ = \left| {{{\matrix{ 3 & {10} & 2 \cr 2 & 1 & { - 2} \cr 2 & { - 3} & 2 \cr } } \over {\left| { - 4\widehat i - 8\widehat j - 8\widehat k} \right|}}} \right| = \left| {{{( - 12) - 10(8) + 2( - 8)} \over {12}}} \right|$

= 9 units

$C:\left( {{{a + 4} \over 3},{{ - 2 + 2b} \over 3},{{4 - 6} \over 3}} \right)$

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

C lies on plane $2x - y + z = 4$

$2\left( {{{a + 4} \over 3}} \right) - \left( {{{2b - 2} \over 3}} \right) - {2 \over 3} = 4$

$ \Rightarrow 2a + 8 - 2b + 2 - 2 = 12$

$ \Rightarrow a - b = 2$ ..... (i)

Now, $OP = \sqrt 5 $

${\left( {{{a + 4} \over 3}} \right)^2} + {\left( {{{2b - 2} \over 3}} \right)^2} + {4 \over 9} = 5$ and using (i)

$a = {{11} \over 5},1$

$ \Rightarrow b = {1 \over 5}, - 1$

as also $ \Rightarrow a = 1,b = - 1$

$P(2, - 1, - 3),c\left( {{5 \over 3},{{ - 4} \over 3},{{ - 2} \over 3}} \right)$

$C{P^2} = {1 \over 9} + {1 \over 9} + {{49} \over 9} = {{17} \over 3}$

${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1} = \lambda $ (say)

& ${{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1} = \mu $ (say)

$\therefore$ $\lambda + 1 = 2\mu + a$ ...... (i)

$2\lambda + 2 = 3\mu - 2$ ..... (ii)

$\lambda - 3 = \mu + 3$ .... (iii)

By (i) & (ii)

$ \Rightarrow 3\mu - 2 = 4\mu + 2a + 2$

$\mu=-2(1+a)$ & $\lambda=5-3a$

Put $\lambda$ & $\mu$ in (iii) we get

$a=-9$

$\mu=16$

$\lambda=22$

$\therefore$ Point of intersection $\equiv(23,46,19)$

Distance from $z=-9$ is 28

${\overrightarrow r _1} = \widehat i - 8\widehat j + 4\widehat k$

${\overrightarrow r _2} = \widehat i + 2\widehat j + 6\widehat k$

$\overrightarrow a = 2\widehat i - 7\widehat j + 5\widehat k$

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

S.D. $ = {{\left| {\matrix{ {{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} & {\matrix{ {\overrightarrow a } & {\overrightarrow b } \cr } } \cr } } \right|} \over {\left| {\overrightarrow a \times \overrightarrow b } \right|}}$

$\left[ {\matrix{ {{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} & {\matrix{ {\overrightarrow a } & {\overrightarrow b } \cr } } \cr } } \right] = \left| {\matrix{ 0 & { - 10} & { - 2} \cr 2 & { - 7} & 5 \cr 2 & 1 & { - 3} \cr } } \right|$

$\therefore$ $10( - 16) - 2(16) = - 192$

$\left| {\left[ {\matrix{ {{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} & {\matrix{ {\overrightarrow a } & {\overrightarrow b } \cr } } \cr } } \right]} \right| = 192$

$\overrightarrow a \times \overrightarrow b = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & { - 7} & 5 \cr 2 & 1 & { - 3} \cr } } \right| = 16\widehat i + 16\widehat j + 16\widehat k$

$\overrightarrow a \times \overrightarrow b = 16\sqrt 3 $

S.D. $ = {{192} \over {16\sqrt 3 }} = 4\sqrt 3 $

$ \mathrm{L}: \frac{\mathrm{x}+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}=\lambda \text { (let) } $

$\left( { - 2,{7 \over 2},{3 \over 2}} \right)$ satisfies the plane $P:2x + my + nz = 4$

$ - 4 + {{7m} \over 2} + {{3n} \over 2} = 4 \Rightarrow 7m + 3n = 16$ ...... (i)

Line joining $A( - 1,\,4,\,3)$ and $\left( { - 2,{7 \over 2},{3 \over 2}} \right)$ is perpendicular to $P:2x + my + nz = 4$

${1 \over 2} = {{{1 \over 2}} \over m} = {{{3 \over 2}} \over n} \Rightarrow m = 1$ & $n = 3$

Plane $P:2x + y + 3z = 4$

Distance of P from $A( - 1,\,4,\,3)$ parallel to the line

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

for point of intersection of P & L

$2(3r - 1) + ( - r + 4) + 3( - 4r + 3) = 4 \Rightarrow r = 1$

Point of intersection : $(2,\,3,\, - 1)$

Required distance $ = \sqrt {{3^2} + {1^2} + {4^2}} = \sqrt {26} $

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

through a point ${\overrightarrow a _1} \equiv (1,2,3)$

parallel to ${\overrightarrow b _1} \equiv (\lambda ,1,2)$

${L_2}:{{x + 26} \over { - 2}} = {{y + 18} \over 3} = {{z + 28} \over \lambda }$

through a point ${\overrightarrow a _2} = ( - 26, - 18, - 28)$

parallel to ${\overrightarrow b _2} = ( - 2,3,1)$

If lines are coplanar then, $({\overrightarrow a _2} - {\overrightarrow a _1})\,.\,{\overrightarrow b _1} \times {\overrightarrow b _2} = 0$

$ \Rightarrow \left| {\matrix{ {27} & {20} & {31} \cr \lambda & 1 & 2 \cr { - 2} & 3 & \lambda \cr } } \right| = 0 \Rightarrow \lambda = 3$

Vector normal to the required plane $\overrightarrow n = {\overrightarrow b _1} \times {\overrightarrow b _2}$

$ \Rightarrow \overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & 1 & 2 \cr { - 2} & 3 & 3 \cr } } \right| = - 3\widehat i - 13\widehat j + 11\widehat k$

Equation of plane

$ \equiv ((x - 1),(y - 2),(z - 3))\,.\,( - 3, - 13,11) = 0$

$ \Rightarrow 3x + 13y - 11z + 4 = 0$

Checking the option gives (0, 4, 5) does not lie on the plane.

Let ${\overrightarrow a _1} = ( - 2,1, - 3)$ and ${\overrightarrow a _2} = ( - 1,2, - 2)$

Vector normal to plane $\overline n = {\overrightarrow a _1} \times {\overrightarrow a _2}$

$\overline n = (4, - 1, - 3)$

Plane through $(2,2, - 2)$ and normal to $\overline n $

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

$ \Rightarrow 4x - y - 3z = 12$

$ \Rightarrow {x \over 3} + {y \over { - 12}} + {z \over { - 4}} = 1$

Intercepts $\alpha$, $\beta$, $\gamma$ are $3, - 12, - 4$

$P = \alpha + \beta + \gamma = - 13$

$V = {1 \over 6} \times 3 \times 12 \times 4 = 24$

Any point on ${x^2} + {y^2} = 1$, $z = 0$ is $p(\cos \theta ,\,\sin \theta ,\,0)$

If foot of perpendicular of p on the plane $2x + 3y + z = 6$ is $(h,k,l)$ then

${{h - \cos \theta } \over 2} = {{k - \sin \theta } \over 3} = {{l - 0} \over 1}$

$ = - \left( {{{2\cos \theta + 3\sin \theta + 0 - 6} \over {{2^2} + {3^2} + {1^2}}}} \right) = r$ (let)

$h = 2r + \cos \theta ,\,k = 3r + \sin \theta ,\,l = r$

Hence, $h - 2l = \cos \theta $ and $k - 3l = \sin \theta $

Hence, ${(h - 2l)^2} + {(k - 3l)^2} = 1$

When $l = 6 - 2h - 3k$

Hence required locus is

${(x - 2(6 - 2x - 3y))^2} + {(y - 3(6 - 2x - 3y))^2} = 1$

$ \Rightarrow {(5x + 6y - 12)^2} + 4{(3x + 5y - 9)^2} = 1,\,z = 6 - 2x - 3y$

$\because$ PR is perpendicular to given line, so

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

$ \Rightarrow a = 7\lambda - 2$

Now,

$\because$ $PR = 2\sqrt 6 $

$ \Rightarrow {( - 5\lambda + 1)^2} + {(3\lambda - 1)^2} + {(\lambda + 1)^2} = 24$

$ \Rightarrow 5{\lambda ^2} - 2\lambda - 3 = 0 \Rightarrow \lambda = 1$ or $ - {3 \over 5}$

$\because$ $a > 0$ so $\lambda = 1$ and $a = 5$

Now $\sum\limits_{i = 1}^3 {{\alpha _i} = 2} $ (Sum of co-ordinate of R) $-$ (Sum of coordinates of P)

$ = 2(7) - 11 = 3$

$a + \sum\limits_{i = 1}^3 {{\alpha _i} = 5 + 3 = 8} $

${P_1}:ax + by + 0z = 3$, normal vector : ${\overrightarrow n _1} = (a,b,0)$

${P_2}:ax + by + cz = 0$, normal vector : ${\overrightarrow n _2} = (a,b,c)$

Vector parallel to the line of intersection $ = {\overrightarrow n _1} \times {\overrightarrow n _2}$

${\overrightarrow n _1} \times {\overrightarrow n _2} = (bc, - ac,0)$

Vector normal to $0\,.\,x + y - z + 2 = 0$ is ${\overrightarrow n _3} = (0,1, - 1)$

Angle between line and plane is 30$^\circ$

$ \Rightarrow \left| {{{0 - ac + 0} \over {\sqrt {{b^2}{c^2} + {c^2}{a^2}} \sqrt 2 }}} \right| = {1 \over 2}$

$ \Rightarrow {a^2} = {b^2}$

Hence, ${\overrightarrow n _1} \times {\overrightarrow n _2} = (ac, - ac,0)$

Direction ratios $ = \left( {{1 \over {\sqrt 2 }}, - {1 \over {\sqrt 2 }},0} \right)$

${P_1}:2x + ky - 5z = 1$

${P_2}:3kx - ky + z = 5$

$\because$ ${P_1}\, \bot \,{P_2} \Rightarrow 6k - {k^2} + 5 = 0$

$ \Rightarrow k = 1,5$

$\because$ $k < 3$

$\therefore$ $k = 1$

${P_1}:2x + y - 5z = 1$

${P_2}:3x - y + z = 5$

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

Positive x-axis intercept = 1

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

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

$\therefore$ $P:7x + y - 4z = 7$

y intercept = 7.

If ${\overrightarrow n _1}$ is a vector normal to the plane determined by $\widehat i$ and $\widehat i + \widehat j$ then

${\overrightarrow n _1} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 0 & 0 \cr 1 & 1 & 0 \cr } } \right| = \widehat k$

If ${\overrightarrow n _2}$ is a vector normal to the plane determined by $\widehat i - \widehat j$ and $\widehat i + \widehat k$ then

${\overline n _2} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & { - 1} & 0 \cr 1 & 0 & 1 \cr } } \right| = - \widehat i - \widehat j + \widehat k$

Vector $\overrightarrow a $ is parallel to ${\overrightarrow n _1} \times {\overrightarrow n _2}$

i.e. $\overrightarrow a $ is parallel to $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 0 & 0 & 1 \cr { - 1} & { - 1} & 1 \cr } } \right| = \widehat i - \widehat j$

Given $\overrightarrow b = \widehat i - 2\widehat j + 2\widehat k$

Cosine of acute angle between

$\overrightarrow a $ and $\overrightarrow b = \left| {{{\overrightarrow a \,.\,\overrightarrow b } \over {|\overrightarrow a |\,.\,|\overrightarrow b |}}} \right| = {1 \over {\sqrt 2 }}$

Obtuse angle between $\overrightarrow a $ and $\overrightarrow b = {{3\pi } \over 4}$

The line $x + y - z = 0 = x - 2y + 3z - 5$ is parallel to the vector

$\overrightarrow b = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 1 & { - 1} \cr 1 & { - 2} & 3 \cr } } \right| = (1,4, - 3)$

Equation of the line through $P(1,2,4)$ and parallel to $\overrightarrow b $

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

Let $N \equiv (\lambda + 1, - 4\lambda + 2, - 3\lambda + 4)$

$\overrightarrow {QN} = (\lambda , - 4\lambda + 4, - 3\lambda - 1)$

$\overrightarrow {QN} $ is perpendicular to $\overrightarrow {b} $

$ \Rightarrow (\lambda , - 4\lambda + 4, - 3\lambda - 1)\,.\,(1,4, - 3) = 0$

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

Hence $\overrightarrow {QN} = \left( {{1 \over 2},2,{{ - 5} \over 2}} \right)$ and $\left| {\overrightarrow {QN} } \right| = \sqrt {{{21} \over 2}} $

First plane, ${P_1} = 2x - 2y + z = 0$, normal vector $ \equiv {\overline n _1} = (2, - 2,1)$

Second plane, ${P_2} \equiv x - y + 2z = 4$, normal vector $ \equiv {\overline n _2} = (1, - 1,2)$

Plane perpendicular to P₁ and P₂ will have normal vector ${\overline n _3}$

Where ${\overline n _3} = \left( {{{\overline n }_1} \times {{\overline n }_2}} \right)$

Hence, ${\overline n _3} = ( - 3, - 3,0)$

Equation of plane E through $P(1, - 1,1)$ and ${\overline n _3}$ as normal vector

$(x - 1,y + 1,z - 1)\,.\,( - 3, - 3,0) = 0$

$ \Rightarrow x + y = 0 \equiv E$

Distance of $PQ(a,a,2)$ from $E = \left| {{{2a} \over {\sqrt 2 }}} \right|$

as given, $\left| {{{2a} \over {\sqrt 2 }}} \right| = 3\sqrt 2 \Rightarrow a = \, \pm \,3$

Hence, $Q \equiv ( \pm \,3, \pm \,3,2)$

Distance $7Q = \sqrt {21} \Rightarrow {(PQ)^2} = 21$

${L_1}:{{x + 7} \over 6} = {{y - 6} \over 7} = {{z - 0} \over 1}$

Any point on it ${\overrightarrow a _1}( - 7,6,0)$ and L₁ is parallel to ${\overrightarrow b _1}( - 6,7,1)$

${L_2}:{{x - 7} \over { - 2}} = {{y - 2} \over 1} = {{z - 6} \over 1}$

Any point on it ${\overrightarrow a _2}(7,2,6)$ and L₂ is parallel to ${\overrightarrow b _2}( - 2,1,1)$

Shortest distance between L₁ and L₂

$ = \left| {{{({{\overrightarrow a }_2} - {{\overrightarrow a }_1})\,.\,({{\overrightarrow b }_1} \times {{\overrightarrow b }_2})} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}}} \right| = \left| {{{( - 14,4, - 6)\,.\,(3,2,4)} \over {\sqrt {9 + 4 + 16} }}} \right|$

$ = 2\sqrt {29} $.

Line PQ is parallel to line ${{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1}$

$\therefore$ DR of PQ = DR of line = <$-$2, 2, 1>

$\therefore$ Equation of line PQ passing through P(3, 2, $-$1) and DR = <$-$2, 2, 1> is

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

Any General point on line PQ = (x₁, y₁, z₁)

$\therefore$ ${{{x_1} - 3} \over { - 2}} = {{{y_1} - 2} \over 2} = {{{z_1} + 1} \over 1} = \lambda $

$ \Rightarrow {x_1} = - 2\lambda + 3$

${y_1} = 2\lambda + 2$

${z_1} = \lambda - 1$

$\therefore$ Point Q = ($-$2$\lambda$ + 3, 2$\lambda$ + 2, $\lambda$ $-$ 1)

Point Q lies on the plane $3x - y + 4z + 1 = 0$. So point Q satisfy the equation.

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

$ \Rightarrow - 6\lambda + 9 - 2\lambda - 2 + 4\lambda - 4 + 1 = 0$

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

$ \Rightarrow \lambda = 1$

$\therefore$ Point Q = ($-$2 $\times$ 1 + 3, 2 $\times$ 1 + 2, 1 $-$ 1)

= (1, 4, 0)

$\therefore$ Distance of the point P(3, 2, $-$1) from the plane = Length of PQ

$ = \sqrt {{{(3 - 1)}^2} + {{(2 - 4)}^2} + {{( - 1 - 0)}^2}} $

$ = \sqrt {{2^2} + {{( - 2)}^2} + {{( - 1)}^2}} $

$ = \sqrt {4 + 4 + 1} $

$ = \sqrt 9 $

$ = 3$

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 $\pi$