3D Geometry

Since the plane bisects the line joining the points (1, 2, 3) and ($-$3, 4, 5) then the plane passes through the midpoint of the line which is :

$\left( {{{1 - 3} \over 2},{{2 + 4} \over 2},{{5 + 3} \over 2}} \right)$  $ \equiv $  $\left( {{{ - 2} \over 2},{6 \over 2},{8 \over 2}} \right)$ $ \equiv $ ($-$1, 3, 4).

As plane cuts the line segment at right angle, so the direction cosines of the normal of the plane are ($-$ 3 $-$ 1, 4 $-$ 2, 5 $-$ 3) = ($-$ 4, 2, 2)

So the equation of the plane is : $-$ 4x + 2y + 2z = $\lambda $

As plane passes through ($-$ 1, 3, 4)

So, $-$ 4($-$ 1) + 2(3) + 2(4) = $\lambda $ $ \Rightarrow $ $\lambda $ = 18

Therefore, equation of plane is : $-$ 4x + 2y + 2z = 18

Now, only ($-$ 3, 2, 1) satiesfies the given plane as

$-$ 4($-$ 3) + 2(2) + 2(1) = 18

If a, b, c are the intercepts of the variable plane on the x,y,z axes respectively, then the equation of the plane is

${x \over a} + {y \over b} + {z \over c} = 1$

And the point of intersection of the planes parallel to the xy, yz and zx planes is $\left( {a,b,c} \right)$.

As the point (3, 2, 1) lies on the variables plane,

so ${3 \over a} + {2 \over b} + {1 \over c} = 1$

Therefore, the required locus is

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

Normal to $3x + 4y + z = 1$    is $3\widehat i + 4\widehat j + \widehat k$

Normal to $5x + 8y + 2z =$ $ - 14$ is $5\widehat i + 8\widehat j + 2\widehat k$

The line of intersection of the planes is perpendicular to both normals, so, direction ratios of the intersection line are directly proportional to the cross product of the normal vectors.

Therefore the direction ratios of the line is $ - \widehat j + 4\widehat k$

Hence the angle between the plane x + y + z + 5 = 0

and the intersection line is ${\sin ^{ - 1}}\left( {{{ - 1 + 4} \over {\sqrt {17} \sqrt 3 }}} \right) = {\sin ^{ - 1}}\left( {\sqrt {{3 \over {17}}} } \right)$

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.

Suppose centroid be (h, k, $l$)

$ \therefore $   x $-$ intp $=$ 3h, y $-$ intp $=$ 3k, z $-$ intp $=$ 3$l$

Equation ${x \over {3h}} + {y \over {3k}} + {z \over {3l}} = 1$

$ \therefore $   Distance from (0, 0, 0)

$\left| {{{ - 1} \over {\sqrt {{1 \over {9{h^2}}} + {1 \over {9{k^2}}} + {1 \over {9{l^2}}}} }}} \right| = 3$

$ \Rightarrow $   ${1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = 1$

Point (3, $-$ 2, $-$ $\lambda $) on p line 2x $-$ 4y + 3z $-$ 2 $=$ 0

$=$ 6 + 8 $-$ 3$\lambda $ $-$ 2 = 0

$=$ 3$\lambda $ $=$ 12

$\lambda $ $=$ 4

Now,

${{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + 4} \over { - 2}} = {k_1}$          . . .(i)

${{x - 1} \over {12}} = {y \over 9} = {z \over 4} = {k_2}$                 . . .(ii)

Point on equation (i) P (k₁ + 3, $-$ k₁ $-$ 2, $-$ 2k₁ $-$ 4)

Point on equation (ii) Q(12k₂ + 1, 9k₂, 4k₂)

k₁ + 3 $=$ 12k₂ + 1 $\left| { - {k_1} - 2 = 9{k_2}} \right|$ $-$ 2k₁ $-$ 4 $=$ 4k₂

k₂  $=$  0

k₁   $=$  $-$ 2

p (1, 0, 0) lie on equation of a line 1

gives shortest distance $=$ 0

Given,

x + 8y + 7z = 0

9x + 2y + 3z = 0

x + y + z = 0

Solving those equations, we get

x = $\lambda $, y = 6$\lambda $, z = -7$\lambda $

This point lies on the plane x + 2y + z = 6

$ \therefore $ $\lambda $ + 2(6$\lambda $) + (-7$\lambda $) = 0

$ \Rightarrow $ $\lambda $ = 1

$ \therefore $ x = 1, y = 6, z = -7

$ \therefore $ a = 1, b = 6, c = -7

So, 2a + b + c = 2(1) + 6 + (-7) = 1

$\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $

$ \Rightarrow $   $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & { - 1} & 1 \cr 1 & 4 & { - 2} \cr } } \right| = \widehat i\left( { - 2} \right) - \widehat j\left( { - 7} \right) + \widehat k\left( {13} \right)$

$\overrightarrow n = - 2\widehat i + 7\widehat j + 13\widehat k$

Now,

        $3x - y + z = 1$

        $x + 4y - 2z = 2$

but   z $=$ 0 & solving the given

        x $=$ 6/13 & y = 5/13

$ \therefore $   required equation of a line is

${{x - 6/13} \over 2} = {{y - 5/13} \over { - 7}} = {z \over { - 13}}$

$\overrightarrow n $ = $\overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $ = $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 6 & 7 & 8 \cr 3 & 5 & 7 \cr } } \right|$

= (9, $-$ 18, 9)

= (1, $-$2, 1)

$ \therefore $   Equation of plane is

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

$ \Rightarrow $   x $-$ 2y + z = 0

foot to z

${{x - 1} \over 1}$ = ${{y + 2} \over { - 2}}$ = ${{z - 1} \over 1}$ = $ - {{\left[ {1 + 4 + 1} \right]} \over 6}$

x = 0,   y = 0,   z = 0

Let the plane be

a(x $-$ 1) + b(y + 1) + c (z + 1) = 0

Normal vector

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

So plane is 5(x $-$ 1) + 7(y + 1) + 3(z + 1) = 0

$ \Rightarrow $   5x + 7y + 3z + 5 = 0

Distance of point (1, 3, $-$ 7) from the plane is

${{5 + 21 - 21 + 5} \over {\sqrt {25 + 49 + 9} }} = {{10} \over {\sqrt {83} }}$

Equation of line PQ is ${{x - 1} \over 1} = {{y + 2} \over 4} = {{z - 3} \over 5}$

Let F be ($\lambda $ + 1, 4$\lambda $ $-$ 2, 5$\lambda $ + 3)


Since F lies on the plane

$ \therefore $   2($\lambda $ + 1) + 3(4$\lambda $ $-$ 2) $-$ 4(5$\lambda $ + 3) + 22 $=$ 0

$ \Rightarrow $   $-$ 6$\lambda $ + 6 = 0 $ \Rightarrow $  $\lambda $ = 1

$ \therefore $   F is (2, 2, 8)

PQ = 2 PF = 2$\sqrt {{1^2} + {4^2} + {5^2}} $ = 2$\sqrt {42} $

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

As lines are coplanar, so

$\left| {\matrix{ {3 - 1} & {2 - 2} & {1 - \left( { - 3} \right)} \cr 1 & 2 & {{\lambda ^2}} \cr 1 & {{\lambda ^2}} & 2 \cr } } \right| $ = 0

$ \Rightarrow $   $\left| {\matrix{ 2 & 0 & 4 \cr 1 & 2 & {{\lambda ^2}} \cr 1 & {{\lambda ^2}} & 2 \cr } } \right| $ = 0

$ \Rightarrow $   2(4 $-$ $\lambda $⁴) + 4($\lambda $² $-$ 2) = 0

$ \Rightarrow $   4 $-$ $\lambda $⁴ + 2$\lambda $² $-$ 4 = 0

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

$ \Rightarrow $   $\lambda $ = 0, $\sqrt 2 , - \sqrt 2 $

$ \therefore $   3 distinct real values are possible.

DR's of AD are

${{\lambda - 1} \over 2} - 2,{{5 + 3} \over 2} - 3,{{\mu + 2} \over 2} - 5$

i.e.  ${{\lambda - 5} \over 2},\,\,1,\,\,{{\mu - 8} \over 2}$

As medium is making equal angles with coordinate axes,

$ \therefore $   ${{\lambda - 5} \over 2} = 1 = {{\mu - 8} \over 2}$

$ \Rightarrow $   $\lambda $ = 7,   $\mu $ = 10

$ \therefore $   $\lambda $³ + $\mu $³ + 5 = 7³ + 10³ + 5 = 1348

Shortest distance between the lines

${{x - {x_1}} \over {{a_1}}} = {{y - {y_1}} \over {{b_1}}} = {{z - {z_1}} \over {{c_1}}}$

and  ${{x - {x_2}} \over {{a_2}}} = {{y - {y_2}} \over {{b_2}}} = {{z - {z_2}} \over {{c_2}}}$   is

   $\left| {{{\left| {\matrix{ {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr {{a_1}} & {{b_1}} & {{c_1}} \cr {{a_2}} & {{b_2}} & {{c_2}} \cr } } \right|} \over {\sqrt {{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2} + {{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2}} }}} \right|$

$ \therefore $   Shortest distance between two given lines are,

   $\left| {{{\left| {\matrix{ { - 2} & 4 & 5 \cr 2 & 2 & 1 \cr { - 1} & 8 & 4 \cr } } \right|} \over {\sqrt {{{\left( {8 - 8} \right)}^2} + {{\left( { - 1 - 8} \right)}^2} + {{\left( {16 + 2} \right)}^2}} }}} \right|$

=   $\left| {{{ - 36 + 90} \over {\sqrt {405} }}} \right|$

=   ${{54} \over {20.1}}$

=   2.68

Equation of plane passing through point (1, 2, 2) is,

a(x $-$ 1) + b(y $-$ 2) + c(z $-$ 2) = 0        . . .(1)

This plane is perpendicular to the plane

x $-$ y + 2z = 3   and  2x $-$ 2y + z + 12 = 0

When two planes,

a₁x + b₁y + c₁z + d₁ = 0

and a₂x + b₂y + c₂z + d₂ = 0

are perpendicular to each other then

a₁a₂ + b₁b₂ + c₁c₂ = 0

So, we can say,

a $ \times $ 1 + b $ \times $ ($-$ 1) + c $ \times $ 2 = 0

$ \Rightarrow $   a $-$ b + 2c = 0           . . . (2)

and, a $ \times $ 2 + b($-$2) + c(1) = 0

$ \Rightarrow $   2a $-$ 2b + c = 0           . . .(3)

Solving (2) and (3)

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

$ \Rightarrow $    a = 3$\lambda $, b = 3$\lambda $, c = 0

Putting the values of a, b, c in equation (1) we get,

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

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

$ \Rightarrow $   3x + 3y $-$ 9 = 0

$ \Rightarrow $   x + y $-$ 3 = 0

$ \therefore $   Distance of point (1, $-$2, 4) from plane x + y $-$ 3 = 0 is,

= $\left| {{{1 - 2 - 3} \over {\sqrt {1 + 1} }}} \right|$ = ${4 \over {\sqrt 2 }}$ = 2$\sqrt 2 $

$e{q^n}\,\,$ of $\,\,PO:\,\,$ ${{x - 1} \over 1} = {{y + 5} \over 1} = {{z - 9} \over 1} = \lambda $

$ \Rightarrow x = \lambda + 1;\,\,y = \lambda - 5;\,\,z = \lambda + 9$

Putting these in $e{q^n}$ of plane :-

$\lambda + 1 - \lambda + 5 + \lambda + 9 = 5$

$ \Rightarrow \lambda = - 10$

$ \Rightarrow O$ is $\left( { - 9, - 15, - 1} \right)$

$ \Rightarrow $ $\,\,\,$ Distance $OP = 10\sqrt 3 .$

Line lies in the plane $ \Rightarrow \left( {3, - 2, - 4} \right)$ lie in the plane

$ \Rightarrow 3\ell - 2m + 4 = 9$ or $3\ell - 2m = 5....\left( 1 \right)$

Also, $\ell ,$ $m, - 1$ are dr's of line perpendicular to plane

and $2, - 1,3$ are dr's of line lying in the plane

$ \Rightarrow 2\ell - m - 3 = 0\,\,\,$ or $\,\,\,2\ell - m = 3....\left( 2 \right)$

Solving $(1)$ and $(2)$ we get $\ell = 1$ and $m=-1$

$ \Rightarrow {\ell ^2} + {m^2} = 2.$

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.

General point on given line $ \equiv P\left( {3r + 2,4r - 1,12r + 2} \right)$

Point $P$ must satisfy equation of plane

$\left( {3r + 2} \right) - \left( {4r - 1} \right) + \left( {12r + 2} \right) = 16$

$11r + 5 = 16$

$r=1$

$P\left( {3 \times 1 + 2,4 \times 1 - 1,12 \times 1 + 2} \right) = P\left( {5,3,14} \right)$

Distance between $P$ and $(1,0,2)$

$D = \sqrt {{{\left( {5 - 1} \right)}^2} + {3^2} + {{\left( {14 - 2} \right)}^2}} = 13$

Equation of the plane containing the lines

$2x - 5y + z = 3$ and $x + y + 4z = 5$ is

$2x - 5y + z - 3 + \lambda \left( {x + y + 4z - 5} \right) = 0$

$ \Rightarrow \left( {2 + \lambda } \right)x + \left( { - 5 + \lambda } \right)y + \left( {1 + 4\lambda } \right)z + $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { - 3 - 5\lambda } \right) = 0....\left( i \right)$

Since the plane $(i)$ parallel to the given plane

$x + 3y + 6z = 1$

$\therefore$ $\,\,\,{{2 + \lambda } \over 1} = {{ - 5 + \lambda } \over 3} = {{1 + 4\lambda } \over 6}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \lambda = - {{11} \over 2}$

Hence equation of the required plane is

$\left( {2 - {{11} \over 2}} \right)x + \left( { - 5 - {{11} \over 2}} \right)y + \left( {1 - {{44} \over 2}} \right)z + $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { - 3 + {{55} \over 2}} \right) = 0$

$ \Rightarrow x + 3y + 6z = 7$

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

${{a - 1} \over 2} = {{b - 3} \over { - 1}} = {{c - 4} \over 1} = \lambda \left( {let} \right)$

$ \Rightarrow a = 2\lambda + 1,\,\,b = 3 - \lambda ,\,\,c = 4 + \lambda $

$p = \left( {{{a + 1} \over 2},{{b + 3} \over 2},{{c + 4} \over 2}} \right)$

$ = \left( {\lambda + 1,{{6 - \lambda } \over 2},{{\lambda + 8} \over 2}} \right)$

$\therefore$ $2\left( {\lambda + 1} \right) - {{6 - \lambda } \over 2} + {{\lambda + 8} \over 2} + 3 = 0$

$3\lambda + 6 = 0 \Rightarrow \lambda = - 2$

$a = - 3,b = 5,c = 2$

Required line is ${{x + 3} \over 3} = {{y - 5} \over 1} = {{z - 2} \over { - 5}}$

Given

$l + m + n = 0$ and ${l^2} = {m^2} + {n^2}$

Now, ${\left( { - m - n} \right)^2} = {m^2} + {n^2}$

$ \Rightarrow mn = 0 \Rightarrow m = 0\,\,$ or $\,\,n = 0$

If $m=0$ then $l=-n$

We know

${l^2} + {m^2} + {n^2} = 1 \Rightarrow n = \pm {1 \over {\sqrt 2 }}$

i.e.$\left( {{l_1},{m_1},{n_1}} \right) = \left( { - {1 \over {\sqrt 2 }},0,{1 \over {\sqrt 2 }}} \right)$

If $n=0$ then $l=-m$

${l^2} + {m^2} + {n^2} = 1\,\,\, \Rightarrow 2{m^2} = 1$

$ \Rightarrow m = \pm {1 \over {\sqrt 2 }}$

Let $m = {1 \over {\sqrt 2 }} \Rightarrow l = - {1 \over {\sqrt 2 }}$

and $n=0$

$\left( {{l_2},{m_2},{n_2}} \right) = \left( { - {1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }},0} \right)$

$\therefore$ $\cos \theta = {1 \over 2} \Rightarrow \theta = {\pi \over 3}$

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 $

Given lines will be coplanar

If $\,\,\,\,\left| {\matrix{ { - 1} & 1 & 1 \cr 1 & 1 & { - k} \cr k & 2 & 1 \cr } } \right| = 0$

$ \Rightarrow - 1\left( {1 + 2k} \right) - \left( {1 + {k^2}} \right) + 1\left( {2 - k} \right) = 0$

$ \Rightarrow k = 0, - 3$

$2x + y + 2z - 8 = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( {Plane\,\,1} \right)$

$2x + y + 2z + {5 \over 2} = 0\,\,\,\,\,\,\,\,\,\,\,...\left( {Plane\,\,2} \right)$

Distance between Plane $1$ and $2$

$ = \left| {{{ - 8 - {5 \over 2}} \over {\sqrt {{2^2} + {1^2} + {2^2}} }}} \right|$

$ = \left| {{{ - 21} \over 6}} \right| = {7 \over 2}$

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

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

Given lines in vector form are

$\overrightarrow r = \left( {\widehat i - \overrightarrow j + \overrightarrow k } \right) + \lambda \left( {2\overrightarrow i + 3\overrightarrow j + 4\overrightarrow j } \right)$

and $\overrightarrow r = \left( {3\widehat i + k\widehat j} \right) + \mu \left( {\widehat i + 2\widehat j + \widehat k} \right)$

These will intersect if shortest distance between them $=0$

i.e.$\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).\overrightarrow {{b_1}} \times {\overrightarrow b _2} = 0$

$ \Rightarrow \left| {\matrix{ {3 - 1} & {k + 1} & { - 1} \cr 2 & 3 & 4 \cr 1 & 2 & 1 \cr } } \right| = 0$

$ \Rightarrow 2\left( { - 5} \right) - \left( {k + 1} \right)\left( { - 2} \right) - 1\left( 1 \right) = 0$

$ \Rightarrow k = 9/2$

Given equation of a plane is

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

So, Equation of parallel plane is given by

$x - 2y + 2z + d = 0$

Now, it is given that distance from origin to the parallel planes is $1.$

$\therefore$ $\left| {{d \over {\sqrt {{1^2} + {2^2} + {2^3}} }}} \right| = 1 \Rightarrow d = \pm 3$

So equation of required plane

$x - 2y + 2z \pm 3 = 0$

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.

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

If $\theta $ be the angle between the given line and plane, then

$\sin \theta = {{1 \times 1 + 2 \times 2 + \lambda \times 3} \over {\sqrt {{1^2} + {2^2} + {\lambda ^2}} .\sqrt {{1^2} + {2^2} + {3^2}} }}$

$ = {{5 + 3\lambda } \over {\sqrt {14} .\sqrt {5 + {\lambda ^2}} }}$

But it is given that

$\theta = {\cos ^{ - 1}}\sqrt {{5 \over {14}}} \Rightarrow \sin \theta = {3 \over {\sqrt {14} }}$

$\therefore$ ${{5 + 3\lambda } \over {\sqrt {14} \sqrt {5 + {\lambda ^2}} }} = {3 \over {\sqrt {14} }} \Rightarrow \lambda = {2 \over 3}$

The directions ratios of the line segment joining points

$A\left( {1,0,7} \right)\,\,$ and $\,\,\,B\left( {1,6,3} \right)$ are $0,6, - 4.$

The direction ratios of the given line are $1,2,3.$

Clearly $1 \times 0 + 2 \times 6 + 3 \times \left( { - 4} \right) = 0$

So, the given line is perpendicular to line $AB.$

Also, the mid point of $A$ and $B$ is $\left( {1,3,5} \right)$ which lies on the given line.

So, the image of $B$ in the given line is $A$, because the given line is the perpendicular bisector of line segment joining points $A$ and $B$, But statement -$2$ is not a correct explanation for statement$-1.$

Direction cosines of the line :

$\ell = \cos {45^ \circ } = {1 \over {\sqrt 2 }},m = \cos {120^ \circ } = {{ - 1} \over 2},\pi = \cos \theta $

where $\theta $ is the angle, which line makes with positive $z$-axis.

Now ${\ell ^2} + {m^2} + {n^2} = 1$

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

$ \Rightarrow \cos \theta = {1 \over 2}\,\,\,\,\left( \theta \right.$ being acute)

$ \Rightarrow 0 = {\pi \over 3}$

$A\left( {3,1,6} \right);\,\,B = \left( {1,3,4} \right)$

Mid-point of $AB = \left( {2,2,5} \right)$ lies on the plane.

and d.r's of $AB=(2,-2,2)$

d.r's of normal to plane $ = \left( {1, - 1,1} \right).$

Direction ratio of $AB$ and normal to the plane are proportional therefore,

$AB$ is perpendicular to the plane

$\therefore$ $A$ is image of $B$

Statement-$2$ is correct but it is not correct explanation.

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

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

As the line ${{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2}$ lie in the plane

$x + 3y - \alpha z + \beta = 0$

$\therefore$ $Pt\left( {2,1, - 2} \right)$ lies on the plane

i.e. $2 + 3 + 2\alpha + \beta = 0$

or $\,\,\,\,2\alpha + \beta + 5 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$

Also normal to plane will be perpendicular to line,

$\therefore$ $\,\,\,3 \times 1 - 5 \times 3 + 2 \times \left( { - \alpha } \right) = 0$

$ \Rightarrow \alpha = - 6$

From equation $(i)$ then, $\beta = 7$

$\therefore$ $\left( {\alpha ,\beta } \right) = \left( { - 6,7} \right)$

Let $P\left( {{x_1},{y_1},{z_1}} \right)$ and $Q\left( {{x_2},{y_2},{z_2}} \right)$ be the initial and final points of the vector whose projections on the three coordinates axes are ${6, - 3,2}$ then

${x_2} - {x_1}, = 6;\,\,{y_2} - {y_1} = - 3;\,\,{z_2} - {z_1} = 2$

So that directions ratios of $\overrightarrow {PQ} $ are ${6, - 3,2}$

$\therefore$ Direction cosines of $\overrightarrow {PQ} $ are

${6 \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }},{{ - 3} \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }},$

$\,\,\,\,\,\,\,\,$ ${2 \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }} = {6 \over 7},{{ - 3} \over 7},{2 \over 7}$