3D Geometry

Point $P(3,2,1)$

Let Line $L_1: \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ and $L_2: \frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$

Direction vector of $L_1:(1,2,1)$

Let the first line be $L_1: \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}=\lambda$.

Any general point $M$ on this line is $M(\lambda+ 1,2 \lambda, \lambda+1)$.

If $M$ is the foot of the perpendicular from $P(3,2,1)$ to $L_1$, the direction ratio of $P M$ is $(\lambda- 2,2 \lambda-2, \lambda)$.

Since $P M \perp L_1$ : Since $P M \perp L_1$ :

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

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

The foot of the perpendicular $M$ is $(2,2,2)$. Since $M$ is the midpoint of $P Q$, where $Q$ is $(a, b, c)$ :

• $\frac{a+3}{2}=2 \Longrightarrow a=1$

• $\frac{b+2}{2}=2 \Longrightarrow b=2$

• $\frac{c+1}{2}=2 \Longrightarrow c=3$

So, $Q=(1,2,3)$.

Find the distance of $Q(1,2,3)$ from the second line

The second line is $L_2: \frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}=\mu$.

Let $A(9,9,5)$ be a point on $L_2$ and $d= 3 \hat{i}+2 \hat{j}-2 \hat{k}$ be its direction vector.

The vector $A Q=(1-9) \hat{i}+(2-9) \hat{j}+(3-5) \hat{k}=-8 \hat{i}-7 \hat{j}-2 \hat{k}$.

The distance $D$ of a point from a line is given by:

Distance of point from line :

$ D=\frac{|\overrightarrow{A Q} \times \vec{d}|}{|\vec{d}|} $

Cross product

$ \begin{gathered} \overrightarrow{A Q} \times \vec{d}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -8 & -7 & -2 \\ 3 & 2 & -2 \end{array}\right| \\ =\hat{i}[(-7)(-2)-(-2)(2)]-\hat{j}[(-8)(-2)-(-2)(3)]+\hat{k}[(-8)(2)-(-7)(3)] \\ =\hat{i}(14+4)-\hat{j}(16+6)+\hat{k}(-16+21) \\ \overrightarrow{A Q} \times \vec{d}=18 \hat{i}-22 \hat{j}+5 \hat{k} \end{gathered} $

Magnitudes

$ \begin{gathered} |\overrightarrow{A Q} \times \vec{d}|=\sqrt{18^2+(-22)^2+5^2}=\sqrt{833} \\ |\vec{d}|=\sqrt{3^2+2^2+(-2)^2}=\sqrt{17} \end{gathered} $

Distance of point from line

$ D=\frac{|\overrightarrow{A Q} \times \vec{d}|}{|\vec{d}|}=\sqrt{\frac{833}{17}}=\sqrt{49}=7 $

Let line $L: \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$, point $P(1,2, a)$.

$ \begin{aligned} & L_1: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}=\lambda_1, \\ & L_2: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}=\lambda_2 . \end{aligned} $

Aany point $Q$ on $L$ is $(t+1,2 t, t+1)$.

Let $Q_1\left(t_1+1,2 t_1, t_1+1\right)$ be point on $L$ and $d_1$ be the distance measured from $P(1,2, a)$ along $L_1$ such that $\left|P Q_1\right|=d_1$.

So, direction ratio of $\overrightarrow{P Q_1}$ and line $L_1$ are proportional.

Direction ratio of $\overrightarrow{P Q_1}$

$ \begin{gathered} =Q_1-P=\left(t_1+1,2 t_1, t_1+1\right)-(1,2, a) =\left(t_1, 2 t_1-2, t_1+1-a\right) \end{gathered} $

Direction ratio of $L_1=(3,4, b)$.

So,

$ \frac{t_1}{3}=\frac{2 t_1-2}{4}=\frac{t_1-a+1}{b} $

Equating

$ \frac{t_1}{3}=\frac{2 t_1-2}{4} \quad \text { and } \quad \frac{t_1}{3}=\frac{t_1-a+1}{b} $

$4 t_1=6 t_1-6 \quad$ and put $t_1=3$

$t_1=3 \quad$ and $\quad \frac{3}{3}=\frac{3-a+1}{b} \Rightarrow a+b=4$…….(1)

Similarly, let $Q_2\left(t_2+1,2 t_2, t_2+1\right)$ be point on $L$ and $d_2$ be the distance measured from $P(1,2, a)$ along $L_2$ such that $\left|P Q_2\right|=d_2$.

so, direction ratio of $\overrightarrow{P Q_2}$ and $d_2$ are proportional.

direction ratio of $\overrightarrow{PQ}$,

$\begin{gathered}=Q_2-P=\left(t_2+1,2 t_2, t_2+1\right)-(1,2, a) \\\\ =\left(t_2, 2 t_2-2, t_2+1-a\right)\end{gathered}$

Direction ratio of $L_2=(1,4, c)$.

$ \frac{t_2}{1}=\frac{2 t_2-2}{4}=\frac{t_2+1-a}{c} $

Equating

$ \frac{t_2}{1}=\frac{2 t_2-2}{4} \quad \text { and } \quad \frac{t_2}{1}=\frac{t_2+1-a}{c} $

$4 t_2=2 t_2-2 \quad$ and put $t_2=-1$

$2 t_2=-2 \quad$ and $\quad \frac{-1}{1}=\frac{-1+1-a}{c}$

$\Rightarrow $ $-1=\frac{-a}{c} \Rightarrow a=c$ ........(2)

Now, calculating

$ d_1^2=\left|P Q_1\right|^2=t_1^2+\left(2 t_1-2\right)^2+\left(t_1+1-a\right)^2, \text { use } t_1=3 $

$\Rightarrow $ $d_1^2=(3)^2+(2 \times 3-2)^2+(3+1-a)^2$

$=9+16+(4-a)^2=25+16+a^2-8 a$

$=a^2-8 a+41$

Similarly,

$ d_2^2=\left|P Q_2\right|^2=t_2^2+\left(2 t_2-2\right)^2+\left(t_2+1-a\right)^2 $

$\Rightarrow $ $d_2^2=(-1)^2+(2 \times(-1)-2)^2+(-1+1-a)^2\left\{\right.$ use $\left.t_2=-1\right\}$

$\Rightarrow $ $ d_2^2=1+16+a^2=17+a^2 $

It is given that $\left|P Q_1\right|=\left|P Q_2\right| \Rightarrow d_1^2=d_2^2$

$\Rightarrow $ $a^2-8 a+41=17+a^2$

$\Rightarrow-8 a+41=17$

$\Rightarrow $ $8 a=41-17=24 \Rightarrow a=3$

from (1) & (2)

$ b=4-a=4-3=1 \quad \text { and } \quad c=a=3 $

So value of $a+b+c=3+1+3=7$

Lines into vector form

$ \begin{gathered} L_1: \vec{r}=-\hat{i}+2 \hat{j}+4 \hat{k}+\lambda(\alpha \hat{i}-\hat{j}-\alpha \hat{k}) \\ L_2: \vec{r}=\hat{j}+\hat{k}+\mu(\alpha \hat{i}+2 \hat{j}+2 \alpha \hat{k}) \end{gathered} $

Lines are not parallel because direction ratios are not proportional

$ \frac{\alpha}{\alpha} \neq \frac{-1}{2} $

So lines are skew lines.

Minimum distance between two skew lines $\vec{r}=\vec{a}+\lambda \vec{b}$ and $\vec{r}=\vec{c}+\mu \vec{d}$ is given as

$ d=\left|\frac{(\vec{a}-\vec{c}) \cdot(\vec{b} \times \vec{d})}{|\vec{b} \times \vec{b}|}\right| $

$\vec{a}=-\hat{i}+2 \hat{j}+4 \hat{k}, \quad \vec{c}=\hat{j}+\hat{k}$

$\vec{b}=\alpha \hat{i}-\hat{j}-\alpha \hat{k}, \quad \vec{d}=\alpha \hat{i}+2 \hat{j}+2 \alpha \hat{k} $

$ \vec{b} \times \vec{d}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2 \alpha\end{array}\right|$

$=\hat{i}(-2 \alpha+2 \alpha)-\hat{j}\left(2 \alpha^2+\alpha^2\right)+\hat{k}(2 \alpha+\alpha)$

$\therefore $ $\vec{b} \times \vec{d}=-3 \alpha^2 \hat{j}+3 \alpha \hat{k}$

$\Rightarrow $ $|\vec{b} \times \vec{d}|=\sqrt{9 \alpha^4+9 \alpha^2}$

$\Rightarrow $ $\vec{a}-\vec{c}=-\hat{i}+\hat{j}+3 \hat{k}$

$\Rightarrow $ $\left|\frac{(\vec{a}-\vec{c}) \cdot(\vec{b} \times \vec{d})}{|\vec{b} \times \vec{d}|}\right|=\sqrt{2}$

$\Rightarrow $ $\left|\frac{(-\hat{i}+\hat{j}+3 \hat{k}) \cdot\left(-3 \alpha^2 \hat{j}+3 \alpha \hat{k}\right)}{\sqrt{9 \alpha^4+9 \alpha^2}}\right|=\sqrt{2}$

$\Rightarrow $ $\left|\frac{-3 \alpha^2+9 \alpha}{3 \sqrt{\alpha^4+\alpha^2}}\right|=\sqrt{2}$

$\Rightarrow $ $\left|\frac{-3 \alpha(\alpha-3)}{3 \alpha \sqrt{\alpha^2+1}}\right|=\sqrt{2}$

$\Rightarrow $ $ \left|\frac{\alpha-3}{\sqrt{1+\alpha^2}}\right|=\sqrt{2} $

squaring

$\Rightarrow $ $ \frac{(\alpha-3)^2}{1+\alpha^2}=2 $

$\Rightarrow $ $\alpha^2+9-6 \alpha=2+2 \alpha^2$

$\Rightarrow $ $ \alpha^2+6 \alpha-7=0 $

This equation has two roots say $\alpha_1$ and $\alpha_2$

So, sum of roots $=-\frac{\text { coefficient of } \alpha}{\text { coefficient of } \alpha^2} $

$\therefore $ $\alpha_1+\alpha_2=-6$

$ \begin{aligned} & n=4 I+m \\ & n(2 m+10 I)+3 I m=0 \\ & \Rightarrow 2(4 I+m)(m+5 I)+3 I m=0 \\ & \Rightarrow 8 I m+2 m^2+10 m l+40 l^2+3 I m=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow 40 l^2+21 m l+2 m^2=0 \\ & \Rightarrow 40 l^2+16 m l+5 m l+2 m^2=0 \\ & \Rightarrow(8 l+m)(5 l+2 m)=0 \end{aligned} $

$ m=-8 I ; 2 m=-5 I $

$ n=-4 i \quad \Rightarrow m=\frac{-5 l}{2} $

$ \begin{aligned} &\begin{aligned} D \cdot R_1 & =\left(l_1, m_1, n_1\right) \\ & =(-1,-8,-4) \\ D R_2 & =\left(l_2, m_2, n_2\right) \\ & =\left(l_2, \frac{-5 l_2}{2}, \frac{3 l_2}{2}\right) \\ =(2,-5,3) & \\ \cos \theta & =\left|\frac{2+40-12}{9 \cdot \sqrt{38}}\right| \\ \cos \theta & =\frac{10}{3 \sqrt{38}} \end{aligned}\\ &\text { i.e. option (1) is correct } \end{aligned} $

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

Any point $(\lambda, 2 \lambda+1,3 \lambda+2)$

Also $B$ lies on $L$

$ \begin{aligned} & \Rightarrow \frac{4}{1}=\frac{\alpha-2}{3} \Rightarrow \alpha=14 \quad \therefore B(4,9,14) \\ & C \equiv\langle 4,9,14\rangle \pm \frac{10}{\sqrt{14}}\langle 1,2,3\rangle \end{aligned} $

Area of $\triangle A B C=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|$

$ \begin{aligned} & =\overrightarrow{A B}=\langle 3,3,11\rangle \\ & \overrightarrow{A C}=\overrightarrow{A B}+\frac{10}{\sqrt{11}}\langle 1,2,3\rangle \end{aligned} $

Cross product of $\overrightarrow{A B}$ to eliminate parallel component

So, area $=\frac{1}{2} \times \frac{10}{\sqrt{14}}(\overrightarrow{A B} \times\langle 1,2,3\rangle)$

$ \overrightarrow{A B} \times\langle 1,2,3\rangle\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 11 \\ 1 & 2 & 3 \end{array}\right| $

$ \begin{aligned} & =\langle-13,2,3\rangle \\ & \text { Area }=\frac{1}{2} \times \frac{10}{\sqrt{14}} \cdot \sqrt{182}=5 \sqrt{13} \end{aligned} $

M is the point of intersection of $L_1 \& L_2$

$ \begin{aligned} & \Rightarrow 2 \lambda-1=2 \mu-1,3 \lambda-1=3 \mu-1,6 \lambda-3=9 \Rightarrow \lambda=2=\mu \\ & M(3,5,9) \end{aligned} $

Now let point P be $(2 K-1,3 K-1,6 K-3)$ on $L_2$

Such that $\mathrm{PM}=7$

$ \begin{aligned} &\begin{aligned} & \Rightarrow \sqrt{(2 K-4)^2+(3 K-6)+(6 K-12)^2}=7 \\ & \Rightarrow 49 K^2+196-196 K=49 \\ & \Rightarrow K^2+4-4 K=1 \Rightarrow K^2-4 K+3=0 \Rightarrow K=1,3 \end{aligned}\\ &\text { So points } \mathrm{P} \& \mathrm{Q} \text { are }(1,2,3) \&(5,8,15) \text { So sum of all co-ordinates of } \mathrm{P} \& \mathrm{Q}=34 \end{aligned} $

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

$ \begin{aligned} & (1+2 t,-1-3 t, t) \\ & 4 t^2+9 t^2+t^2=16 \times 14 \\ & t= \pm 4 \\ & \text { Points are }(9,-13,4),(-7,11,-4) \\ & (\alpha, \beta, \gamma)=(-7,11,-4) \\ & \bar{a}=(-7,11,-4), \bar{c}=(-5,10,3) \\ & \bar{c}-\bar{a}=(2,-1,7) \\ & b \times d=(-1,5,-3) \\ & S . D=\left|\frac{-2-5-21}{\sqrt{1+25+9}}\right|=\frac{28}{\sqrt{35}} \end{aligned} $

$ \begin{aligned} & P(1,2, a), Q(5, b, c) \frac{x-6}{3}=\frac{y-7}{2}=\frac{7-z}{2} \\ & P Q \perp \text { line } \\ & P Q(4, b-2, c-a) \\ & 3(4)+2(b-2)-2(c-a)=0 \\ & a-c=-12 \\ & a=3, c=15, b=8 \\ & a^2+b^2+c^2=9+64+225=298 \end{aligned} $

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

∴ General point R on line is $r(\lambda-3, \lambda+5, \lambda+2)$

$ \begin{aligned} &\overrightarrow{P R}=(\lambda-1, \lambda+5-r, \lambda+1)\\ &\text { Now } \overrightarrow{P R}=\vec{d}=0\\ &\Rightarrow(\lambda-1) 1+(\lambda+5-r) 1+(\lambda+1) 1=0 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow 3 \lambda+5=0 \\ & \Rightarrow \lambda=\frac{r-5}{3} \\ & \therefore R \equiv\left(\frac{r-5}{3}-3, \frac{r-5}{3}+5, \frac{r-5}{3}+2\right) \\ & \therefore R \equiv\left(\frac{r-14}{3}, \frac{r+10}{3}, \frac{r+1}{3}\right) \\ & \text { Now } P R=\sqrt{\frac{14}{3}} \Rightarrow(P R)^2=\frac{14}{3} \\ & \Rightarrow\left(\frac{r-14}{3}+2\right)+\left(\frac{r+10}{3}-r\right)^2+\left(\frac{r+1}{3}-1\right)^2=\frac{14}{3} \\ & \Rightarrow \frac{(r-8)^2}{9}+\frac{(10-2 r)^2}{9}+\frac{(r-2)^2}{9}=\frac{14}{3} \\ & \Rightarrow\left(r^2-16 r+64\right)+\left(100+4 r^2-40 r\right)+\left(r^2-4 r+4\right)=42 \\ & \Rightarrow 6 r^2-60 r+126=0 \\ & \Rightarrow r^2-10 r+21=0 \\ & \Rightarrow r=3,7 \end{aligned}\\ &\text { Sum of possible value of } r \text { is }=10 \end{aligned} $

$ \begin{aligned} & L_1: \frac{x-2}{-3}=\frac{y-6}{2}=\frac{z-7}{4}=t \\ & L_2: \frac{x-4}{2}=\frac{y-3}{1}=\frac{z-5}{3}=s \end{aligned} $

Any point on $L_1$ is $C=(-3 t+2,2 t+6,4 t+7)$

Any point on $L_2$ is $D=(2 s+4, s+3,3 s+5)$

d.r's of $\overrightarrow{C D}=(2 s+3 t+2, s-2 t-3,3 s-4 t-2)$

CD is parallel to $-3 i+5 j+16 k$

$ \begin{aligned} & \frac{2 s+3 t+2}{-3}=\frac{s-2 t-3}{5}=\frac{3 s-4 t-2}{16} \\ & 13 s+9 t+1=0 \ldots .(i) \\ & 41 s+36 t+26=0 \ldots .(2) \end{aligned} $

Solve (i) \& (2) $s=2 ; t=-3$

$ \begin{aligned} & C=(11,0,-5) \\ & D=(8,5,11) \\ & C D^2=290 \end{aligned} $

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

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

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

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

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

So,

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

Simplifying,

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

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

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

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

Using the section formula,

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

Hence,

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

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

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

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

Substituting given values,

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

Therefore, Option (D) is correct.

The equation of line $ RS $ is

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

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

The direction ratios of $ PH $ are

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

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

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

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

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

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

Therefore, Option (A) is correct.

The plane P contains the line

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

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

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

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

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

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

On simplifying,

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

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

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

On simplifying,

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

Hence, equation of plane $P$ is:

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

Option A is correct.

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

$7x - 5y + z = d$

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

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

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

Therefore, equation of $P_1$ is

$7x - 5y + z = 20$

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

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

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

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

Option B is incorrect.

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

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

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

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

Option C is incorrect.

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

Let the normal vectors be:

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

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

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

Substitute the values:

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

Hence,

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

Option D is correct.

$\begin{aligned} &\begin{aligned} & \overrightarrow{\mathrm{p}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}, \overrightarrow{\mathrm{q}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} \\ & \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}\right|=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}} \\ & \mathrm{~A} \equiv(1,2,3) \mathrm{B} \equiv(\lambda, 4,5) \\ & \text { Shortest Distance }=\left|\frac{\overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}\right| \\ & \frac{1}{\sqrt{6}}=\left|\frac{(\lambda-1) \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})}{\sqrt{6}}\right| \\ & \Rightarrow|-\lambda+1+4-2|=1 \Rightarrow|\lambda-3|=1 \\ & \Rightarrow \lambda=3 \pm 1=4,2 \end{aligned}\\ &\text { Radius of circle passing through points }\\ &\begin{aligned} & (0,0),(4,2) \&(2,4) \\ & =\frac{a b c}{4 \Delta}=\frac{\sqrt{20} \times \sqrt{20} \times \sqrt{8}}{4 \times \frac{1}{2}\left|\begin{array}{lll} 1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4 \end{array}\right|}=\frac{20 \times 2 \sqrt{2}}{2 \times 12} \\ & =\frac{5 \sqrt{2}}{3} \end{aligned} \end{aligned}$

The line we need to find should be orthogonal to two given lines, meaning it will be parallel to the cross product of their direction vectors:

$ (\hat{i} + a \hat{j} + b \hat{k}) \times (-b \hat{i} + a \hat{j} + 5 \hat{k}) $

Calculating the cross product results in the vector:

$ \hat{i}(5a - ab) - \hat{j}(b^2 + 5) + \hat{k}(a + ab) $

This shows that the direction ratios of the required line are proportional to:

$ \alpha(5a - ab), - (b^2 + 5), (a + ab) $

Since the line is also expressed as $ \frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4} $, the direction ratios based on this equation are:

$ \alpha(-2), d, -4 $

Thus, we have the equation:

$ \frac{5a - ab}{-2} = \frac{- (b^2 + 5)}{d} = \frac{a + ab}{-4} $

The point $\left(0, -\frac{1}{2}, 0 \right)$ lies on the line, which implies:

$ \frac{0-1}{-2} = \frac{-\frac{1}{2} + 4}{d} = \frac{0-c}{-4} $

Simplifying these gives $ d = 7 $ and $ c = 2 $.

Substitute into the main equation:

$ \frac{5a - ab}{-2} = \frac{a + ab}{-4} $

$ \frac{-b^2 - 5}{7} = \frac{a + ab}{-4} $

Solving these yields:

$ \begin{array}{c|c} -20a + 4ab = -2a - 2ab & 4b^2 + 20 = 70 + 7ab \\ 18a = 6ab & 36 + 20 = 70 + 21a \\ b = 3 & 56 = 28a \Rightarrow a = 2 \end{array} $

Substituting these into the variables, we find:

$ a + b + c + d = 2 + 3 + 2 + 7 = 14 $

$\begin{aligned} & \mathrm{L}_1: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-0}{2} \\ & \text { Let } \mathrm{Q}(\lambda+1, \lambda+2, \lambda) \\ & \overrightarrow{\mathrm{PQ}}=(\lambda-4, \lambda-1, \lambda+3) \\ & \overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{~m}}=0 \end{aligned}$

$\begin{aligned} &\begin{aligned} & \Rightarrow \lambda-4+\lambda+1, \lambda+3=0 \\ & \Rightarrow 3 \lambda=0 \\ & \Rightarrow \lambda=0 \\ & \Rightarrow \mathrm{Q}(1,2,0) \\ & \mathrm{L}_2: \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-0}{1}=\frac{\mathrm{z}-1}{2} \end{aligned}\\ &\text { Let } \mathrm{R}(\mu+2, \mu, \mu+1) \overrightarrow{\mathrm{PR}}=(\mu-3, \mu-1, \mu+4)\\ &\overrightarrow{\mathrm{PR}} \cdot \overrightarrow{\mathrm{n}}=0\\ &\mu-3+\mu-1+\mu+4=0\\ &\neq \mu=0\\ &\mathrm{R}(2,0,1) \end{aligned}$

$\begin{aligned} & \text { Area of } \triangle \mathrm{PQR}(\mathrm{~A})=\frac{1}{2}|\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PR}}| \\ & \mathrm{A}=\frac{1}{2}|(-4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \times(-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}})| \\ & \mathrm{A}=\frac{1}{2}|7(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})| \\ & \left|\begin{array}{lrr} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -4 & 1 & 3 \\ -3 & -1 & 4 \end{array}\right| \\ & =7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\ & 4 \mathrm{~A}^2=49 \times 3=147 \end{aligned}$

$\begin{aligned} & L: \frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c} \\ & L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda \text { (say) } \end{aligned}$

Any point on $L_1$ be $A(2 \lambda+1,3 \lambda+1,4 \lambda+1)$

$L_2: \frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}=\mu \text { (say) }$

Any point on $L_2$ be $B(\mu+3,2 \mu-4, \mu)$

$D$ of $L$ be : $<2 \lambda, 3 \lambda-2,4 \lambda>$ or $<\mu+2,2 \mu+3$, $\mu-1>$

Now $\frac{2 \lambda}{\mu+2}=\frac{3 \lambda-7}{2 \mu+3}=\frac{4 \lambda}{\mu-1}$

$\begin{aligned} &\begin{aligned} & \Rightarrow \lambda=\frac{-6}{5} \quad \mu=-5 \\ & \therefore< a, b, c >\equiv\langle-3,-7,-6 >\text { or }<3,7,6> \\ & \therefore \quad L: \frac{x-1}{3}=\frac{y-1}{7}=\frac{z-1}{6} \end{aligned}\\ &(7,15,13) \text { lies on the line. } \end{aligned}$

$\begin{aligned} & d=\left|\frac{(\vec{a}-\vec{b}) \cdot \vec{p}_1 \times \vec{p}_2}{\left|\vec{p}_1 \times \vec{p}_2\right|}\right|=\frac{5}{\sqrt{6}} \\ & \vec{a}-\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k} \\ & \overrightarrow{p_1} \times \overrightarrow{p_2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & \alpha & 1 \end{array}\right| \\ & \quad=\hat{i}(3-4 \alpha)-\hat{j}(-2)+\hat{k}(2 \alpha-3) \\ & (\vec{a}-\vec{b}) \cdot\left|\overrightarrow{p_1} \times \overrightarrow{p_2}\right|=3-4 \alpha+4-4 \alpha+6 \\ & =13-8 \alpha \end{aligned}$

$\begin{aligned} & \left|\frac{13-8 \alpha}{\sqrt{(3-4 \alpha)^2+4+(2 \alpha-3)^2}}\right|=\frac{5}{\sqrt{6}} \\ & \left|\frac{13-8 \alpha}{\sqrt{20 \alpha^2-36 \alpha+22}}\right|=\frac{5}{\sqrt{6}} \\ & =6(13-8 \alpha)^2=25\left(20 \alpha^2-36 \alpha+22\right) \\ & =116 \alpha^2+348 \alpha-464=0 \\ & \text { Sum of roots }=-3 \end{aligned}$

$L_1: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1} ; L_2: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$

$\begin{aligned} & \cos \theta=\left|\frac{3+0-5}{\sqrt{2} \times \sqrt{50}}\right| \\ & =\frac{2}{10}=\frac{1}{5} \\ & \therefore \sin \theta=\frac{2 \sqrt{6}}{5} \\ & \text { Area }=\frac{1}{2} a b \sin \theta \\ & =\frac{1}{2} \times \sqrt{15} \times \sqrt{15} \times \frac{2 \sqrt{6}}{5} \\ & =3 \sqrt{6} \\ & (\text { Area })^2=9 \times 6=54 \end{aligned}$

$\begin{aligned} & \frac{x+1}{3}=\frac{y}{4}=\frac{z}{5} ;(p \hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k}) \\ & d=\left|\frac{(\vec{a}-\vec{b}) \cdot\left(\vec{p}_1 \times \vec{p}_2\right)}{\left|\vec{p}_1 \times \vec{p}_2\right|}\right|=\frac{1}{\sqrt{6}} \\ & \vec{a}-\vec{b}=(p+1) \hat{i}+2 \hat{j}+\hat{k} \\ & \vec{p}_1 \times \vec{p}_2=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{array}\right|=\hat{i}-2 \hat{j}+\hat{k} \\ & \frac{1}{\sqrt{6}}=\left|\frac{(p+1)-4+1}{\sqrt{6}}\right| \\ & =|p-2|=1 \Rightarrow p=3,1 \\ & a=1, b=3 \\ & \frac{x^2}{1}+\frac{y^2}{9}=1 \end{aligned}$

Length of $L R=\frac{2 a^2}{b}=\frac{2}{3}$

$\begin{aligned} & L_1: \frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1} \\ & L_2: \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4} \end{aligned}$

$\begin{aligned} & a_1:(3, \alpha, 3) \quad a^2=(-3,-7, \beta) \\ & \vec{b}_1=3 \hat{i}-\hat{j}+\hat{k} \quad \vec{b}_2=-3 \hat{i}+2 \hat{j}+4 \hat{k} \end{aligned}$

$d=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\vec{b}_1 \times \vec{b}_2\right|}=3 \sqrt{30}$

$=\frac{\left|\begin{array}{ccc}6 & \alpha+7 & 3-\beta \\ 3 & -1 & 1 \\ -3 & 2 & 4\end{array}\right|}{\sqrt{6^2+15^2+3^2}}=3 \sqrt{30}$

$\begin{aligned} \Rightarrow & |-15 \alpha-3 \beta-132|=270 \\ & |5 \alpha+\beta+44|=90 \\ \Rightarrow & 5 \alpha+\beta=90-44=46 \end{aligned}$

$L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$

Point $A(3 \lambda+6,2 \lambda+7,7-2 \lambda)$

$B(3 \mu+6,2 \mu+7,7-2 \mu)$

Let $P(3 k+6,2 k+7,7-2 k)$ be foot of perpendicular from $P^{\prime}(1,2,3)$

$\begin{aligned} & \therefore \quad P P^{\prime}\langle 3,2,-2\rangle=0 \\ & 3(3 k+5)+(2 k+5) 2+(4-2 k)(-2)=0 \\ & 9 k+15+4 k+10-8+4 k=0 \\ & 17 k+17=0 \\ & \Rightarrow k=-1 \qquad \therefore \quad P(3,5,9)\\ & |\overrightarrow{A P}|=2 \sqrt{17} \end{aligned}$

$\begin{aligned} & \Rightarrow(3 \lambda+3)^2+(2 \lambda+2)^2+(-2-2 \lambda)^2=17 \times 4 \\ & =17(\lambda+1)^2=17 \times 4 \\ & \Rightarrow \lambda+1= \pm 2 \Rightarrow \lambda=1 \text { or } \lambda=-3 \\ & \therefore A(9,9,5), B(-3,1,13) \\ & \overrightarrow{O A} \cdot \overrightarrow{O B}=(9 \hat{i}+9 \hat{j}+5 \hat{k}) \cdot(-3 \hat{i}+\hat{j}+13 \hat{k}) \\ & =-27+9+65=47 \end{aligned}$

Given:

Each of the angles $\beta$ and $\gamma$ is half of the angle that the line makes with the positive $x$-axis, i.e., $\beta = \gamma = \frac{\alpha}{2}$.

The equation for the direction cosines of angles a line makes with the coordinate axes is given by:

$ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 $

Since $\beta = \gamma$, we substitute to get:

$ \cos^2 \alpha + 2 \cos^2 \beta = 1 $

Substitute $\cos \beta = \cos \gamma$:

$ \cos \alpha = 2 \cos^2 \beta - 1 $

Replacing back into the equation:

$ (2 \cos^2 \beta - 1)^2 + 2 \cos^2 \beta = 1 $

Simplify:

$ (2 \cos^2 \beta - 1)(2 \cos^2 \beta + 1) = 0 $

Solving for $\cos^2 \beta$, we have:

$ 2 \cos^2 \beta - 1 = 0 \quad \text{or} \quad 2 \cos^2 \beta + 1 = 0 $

The latter gives no real solutions, thus:

$ \cos^2 \beta = \frac{1}{2} $

Therefore, $\beta = \frac{\pi}{4}$ or $\beta = \frac{\pi}{2}$.

Thus, the sum of all possible values of $\beta$ is:

$ \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} $

Equation of line passing through $P(7,10,11)$ along the line $\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}$ is

$\frac{x-7}{2}=\frac{y-10}{3}=\frac{z-11}{6}=\lambda$

Let the point on the line is

$Q(2 \lambda+7,3 \lambda+10,6 \lambda+11)$

$Q$ lies on line $\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$

$\begin{aligned} & 3 \lambda+10=4 \Rightarrow \lambda=-2 \\ & \therefore \quad Q(3,4,-1) \\ & P Q=\sqrt{16+36+144}=14 \end{aligned}$

$\begin{aligned} & A=(\alpha, \beta, 4)=(2 \lambda+1,3 \lambda+2,4 \lambda+3) \\ & \text { and } B=(a, b, c)=(\mu+6, \mu,-\mu+4) \\ & \because \frac{\mu+2}{2 \lambda-3}=\frac{\mu-1}{3 \lambda+1}=\frac{-\mu+4}{4 \lambda+3} \\ & \therefore \quad \lambda \mu+8 \lambda+4 \mu=1 \quad \ldots(1)\\ & \text { and } 7 \lambda \mu-16 \lambda+4 \mu=7 \quad \ldots(2) \end{aligned}$

from equation (1) and (2)

$\lambda=-1 \text { and } \mu=3$

Or $\lambda=-\frac{1}{3}$ and $\mu=1$

By taking, $\lambda=-1$ and $\mu=3$, we get

$\begin{aligned} & \therefore \quad A(\alpha, \beta, 4)=(-1,-1,-1) \\ & \text { and } B(a, b, c)=(9,3,1) \end{aligned}$

$\therefore\left|\begin{array}{ccc} 1 & 0 & 1 \\ -1 & -1 & -1 \\ 9 & 3 & 1 \end{array}\right|=8$

$\begin{aligned} & 3=\frac{|[(\lambda-1) \hat{i}+3 \hat{j}+3 \hat{k}] .(\hat{k} \times \hat{j})|}{|\hat{k} \times \hat{j}|}=|\lambda-1|=3 \\ & \Rightarrow \lambda=4 \text { or }-2, \lambda_2<\lambda_1 \Rightarrow \lambda_2=-2 \\ & \lambda_1=4 \end{aligned}$

$\begin{aligned} & (3,-4,4-k) \perp \hat{k} \\ & 4-k=0 \Rightarrow k=4 \\ & \Rightarrow \text { Distance between }(4,-2,7) \&(1,2,7) \\ & \Rightarrow \sqrt{3^2+4^2}=5 \end{aligned}$

Let $X$ be mid point of $P$ and $Q$, which would be also feet of perpendicular.

Let $X$ divides $A$ and $B$ in $\lambda: 1, \lambda \neq-1$

$\begin{aligned} &X=\left(\frac{3 \lambda+4}{\lambda+1}, \frac{5 \lambda+7}{\lambda+1}, \frac{3 \lambda+1}{\lambda+1}\right)\\ &\text { Now } P X \perp A B \Rightarrow \overrightarrow{P X} \cdot \overrightarrow{A B}=0\\ &\begin{aligned} &\left(\frac{3 \lambda+4}{\lambda+1}-1\right) \cdot(4-3)+\left(\frac{5 \lambda+7}{\lambda+1}-0\right)(7-5) \\ &+\left(\frac{3 \lambda+1}{\lambda+1}-3\right) \cdot(1-3)=0 \end{aligned} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \frac{2 \lambda+3}{\lambda+1}+\frac{10 \lambda+14}{\lambda+1}+\frac{4}{\lambda+1}=0 \\ & \Rightarrow \frac{12 \lambda+21}{\lambda+1}=0 \Rightarrow \lambda=\frac{-7}{4} \\ & X=\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right) \end{aligned}\\ &X \text { is mid point of } P Q\\ &\begin{aligned} & Q \equiv\left(2 \cdot \frac{5}{3}-1,2 \cdot \frac{7}{3}-0,2 \cdot \frac{17}{3}-3\right) \equiv(\alpha, \beta, \gamma) \\ & \Rightarrow \quad \alpha+\beta+\gamma=\frac{2(5+7+17)}{3}-4=\frac{58}{3}-4=\frac{46}{3} \end{aligned} \end{aligned}$

The line $ L_1 $ is parallel to the vector $ \overrightarrow{\mathrm{a}} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k} $ and passes through the point $ (7, 6, 2) $. The line $ L_2 $ is parallel to the vector $ \overrightarrow{\mathrm{b}} = 2 \hat{i} + \hat{j} + 3 \hat{k} $ and passes through the point $ (5, 3, 4) $.

Equations of the Lines:

Equation of $ L_1 $:

$ \mathbf{r} = 7 \hat{i} + 6 \hat{j} + 2 \hat{k} + \lambda(-3 \hat{i} + 2 \hat{j} + 4 \hat{k}) $

Equation of $ L_2 $:

$ \mathbf{r} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k} + \mu(2 \hat{i} + \hat{j} + 3 \hat{k}) $

Shortest Distance Between $ L_1 $ and $ L_2 $:

The formula for the shortest distance between two skew lines is given by:

$ \text{Distance} = \left|\frac{\left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right| $

Calculations:

Vector between the points:

$ \vec{a}_2 - \vec{a}_1 = (5 \hat{i} + 3 \hat{j} + 4 \hat{k}) - (7 \hat{i} + 6 \hat{j} + 2 \hat{k}) = -2 \hat{i} - 3 \hat{j} + 2 \hat{k} $

Cross product of direction vectors:

$ \vec{b}_1 = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}, \quad \vec{b}_2 = 2 \hat{i} + \hat{j} + 3 \hat{k} $

$ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 4 \\ 2 & 1 & 3 \end{vmatrix} = 2 \hat{i} + 17 \hat{j} - 7 \hat{k} $

Magnitude of the cross product:

$ \left|\vec{b}_1 \times \vec{b}_2\right| = \sqrt{2^2 + 17^2 + (-7)^2} = \sqrt{342} $

Dot product:

$ \left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \times \vec{b}_2\right) = (-2 \hat{i} - 3 \hat{j} + 2 \hat{k}) \cdot (2 \hat{i} + 17 \hat{j} - 7 \hat{k}) $

$ = (-2)(2) + (-3)(17) + (2)(-7) = -4 - 51 - 14 = -69 $

Shortest Distance:

$ \text{Distance} = \left|\frac{-69}{\sqrt{342}}\right| = \frac{23}{\sqrt{38}} $

$\begin{aligned} & H:(5 \lambda-3,2 \lambda+1,3 \lambda-4) \\ & <\mathrm{DR} \text { of } \mathrm{PH}> \\ & <5 \lambda-3,2 \lambda-1,3 \lambda-7> \\ & \overrightarrow{P H} \cdot \overrightarrow{Q R}=0 \\ & \Rightarrow(5 \lambda-3) 5+(2 \lambda-1) 2+(3 \lambda-7) 3=0 \\ & \Rightarrow 25 \lambda-15+4 \lambda-2+9 \lambda-21=0 \\ & \Rightarrow 38 \lambda=38 \\ & \Rightarrow \lambda=1 \\ & \therefore \quad H(2,3,-1) \\ & P H=\sqrt{4+1+16}=\sqrt{21} \\ & \therefore \quad \text { Area }=\frac{1}{2} \times P H \text { QR } \\ & \quad=\frac{1}{2} \times \sqrt{21} \times 5=\frac{5 \sqrt{21}}{2}=\frac{m}{n} \\ & 2 m-5 \sqrt{21} n=0 \end{aligned}$

$\begin{aligned} & \operatorname{ar}(\triangle A B C)=5 \\ & \frac{1}{2} \times b c=5 \\ & \Rightarrow \quad b c=10 \\ & \operatorname{ar}(\triangle A C D)=6 \\ & \frac{1}{2} \times c d=6 \\ & \Rightarrow \quad c d=12 \\ & \operatorname{ar}(\triangle A B D)=7 \\ & b d=14 \end{aligned}$

$\begin{aligned} & \operatorname{area}(\triangle B C D)=\frac{1}{2}|\overrightarrow{B C} \times \overrightarrow{B D}| \\ & \overrightarrow{B C}=<-b, c, 0> \\ & \overrightarrow{B D}=<-b, 0, d> \\ & |\overrightarrow{B C} \times \overrightarrow{B D}|=\sqrt{c^2 d^2+b^2 d^2+b^2 c^2} \\ & =\sqrt{12^2+14^2+10^2} \\ & =\sqrt{440} \\ & \operatorname{ar}(\triangle B C D)=\sqrt{110} \end{aligned}$

To find the distance between the points $ P $ and $ Q $, let's consider the following steps:

Determine the direction vector of line $ L $:

For the line $ L $ to be perpendicular to both given lines, we find the direction vector of $ L $ using the cross product of the direction vectors of the given lines.

The first line has direction vector $ \langle 2, 1, -2 \rangle $ and the second line has direction vector $ \langle 1, 3, 4 \rangle $. The cross product is:

$ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix} = 10\hat{i} - 10\hat{j} + 5\hat{k} = 5(2\hat{i} - 2\hat{j} + \hat{k}) $

Thus, the direction vector of line $ L $ is $ \langle 2, -2, 1 \rangle $.

Equation of line $ L $:

Since $ L $ passes through point $ P(2, -1, 3) $ with direction vector $ \langle 2, -2, 1 \rangle $, its parametric form is:

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

Find the intersection with the $ yz $-plane:

On the $ yz $-plane, $ x = 0 $. Substitute into the line equation to find $ \lambda $:

$ 2\lambda + 2 = 0 \implies \lambda = -1 $

Substitute $ \lambda = -1 $ back into the parametric equations to find $ Q $:

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

Therefore, the intersection point $ Q $ is $ (0, 1, 2) $.

Calculate the distance between $ P $ and $ Q $:

$ \text{d}(P, Q) = \sqrt{(2-0)^2 + (-1-1)^2 + (3-2)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 $

Thus, the distance between points $ P $ and $ Q $ is 3.

$\begin{aligned} & \mathrm{L}: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}-2}{2}=\mu \\ & \mathrm{P}(\mu+1,-\mu-1,2 \mu+2) \\ & \overrightarrow{\mathrm{AP}} \cdot \overrightarrow{\mathrm{~d}}=0 \Rightarrow(\mu,-\mu-3,2 \mu) \cdot(1,-1,2)=0 \\ & \Rightarrow \mu+\mu+3+4 \mu=0 \Rightarrow \mu=-\frac{1}{2} \\ & \therefore \mathrm{P}\left(\frac{-1}{2}+1,+\frac{1}{2}-1,2\left(\frac{-1}{2}\right)+2\right) \\ & \mathrm{P}\left(\frac{1}{2}, \frac{-1}{2}, 1\right) \end{aligned}$

$\begin{array}{c|c|c} \mu+1=-1+\lambda & -\mu-1=1-\lambda & 2 \mu+2=-2+\lambda \\ \mu=\lambda-2 & \mu=\lambda-2 & \downarrow \end{array}$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\begin{aligned} & 2(\lambda-2)+2=-2+\lambda \\ & 2 \lambda-4+2=-2+\lambda \end{aligned}$

$\begin{aligned} & \therefore \quad \mu=-2, \lambda=0 \\ & \therefore \quad \mathrm{Q} \equiv(-1,1-2) \\ & \mathrm{P}\left(\frac{1}{2}, \frac{-1}{2}, 1\right) \text { and } \mathrm{Q}(-1,1,-2) \\ & \mathrm{PQ}=\sqrt{\left(\frac{1}{2}+1\right)^2+\left(\frac{-1}{2}-1\right)^2+(1+2)^2} \\ & =\sqrt{\frac{9}{4}+\frac{9}{4}+9}=\sqrt{\frac{54}{4}} \\ & \therefore \quad 2(\mathrm{PQ})^2=2\left(\frac{54}{4}\right)=27 \end{aligned}$

Step 1. Identify the Given Lines

The line

$ \mathrm{L}_1:\; \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2} $

passes through the point

$ P_1=(1,2,1) $

with direction vector

$ \mathbf{u}=(1,-1,2). $

Similarly, the line

$ \mathrm{L}_2:\; \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1} $

passes through the point

$ P_2=(-1,2,0) $

with direction vector

$ \mathbf{v}=(-1,2,1). $

Step 2. Determine the Direction of $ L_3 $

Since $ L_3 $ is perpendicular to both $ L_1 $ and $ L_2 $, its direction vector must be parallel to the cross product of $ \mathbf{u} $ and $ \mathbf{v} $. Compute:

$ \mathbf{d} = \mathbf{u}\times\mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{vmatrix}. $

Using the determinant, we obtain:

$ \begin{aligned} d_x &= (-1)(1) - (2)(2) = -1-4=-5,\\[1mm] d_y &= (2)(-1) - (1)(1) = -2-1=-3,\\[1mm] d_z &= (1)(2) - (-1)(-1) = 2-1=1. \end{aligned} $

Thus, a valid direction vector is

$ \mathbf{d}=(-5,-3,1). $

Step 3. Express $ L_3 $ in Terms of Its Intersection with $ L_1 $

Since $ L_3 $ intersects $ L_1 $, let the intersection point (on $ L_1 $) be expressed using a parameter $ t $ as:

$ A(t) = (1+t,\, 2-t,\, 1+2t). $

Since $ L_3 $ passes through $ A(t) $ and also through the given point

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

its parametric form can be written as:

$ (\alpha,\beta,\gamma)=A(t) + s\,\mathbf{d} = (1+t,\,2-t,\,1+2t) + s(-5,-3,1) $

for some parameters $ s $ and $ t $.

Step 4. Express $\alpha,\beta,\gamma$ in Terms of $t$ and $s$

Comparing coordinates, we have:

$ \begin{cases} \alpha = 1 + t - 5s,\\[1mm] \beta = 2 - t - 3s,\\[1mm] \gamma = 1 + 2t + s. \end{cases} $

Step 5. Compute $5\alpha - 11\beta - 8\gamma$

Substitute the expressions for $\alpha$, $\beta$, and $\gamma$:

$ \begin{aligned} 5\alpha - 11\beta - 8\gamma &= 5(1+t-5s) - 11(2-t-3s) - 8(1+2t+s)\\[1mm] &= (5 + 5t - 25s) - (22 - 11t - 33s) - (8 + 16t + 8s)\\[1mm] &= \left(5 - 22 - 8\right) + \left(5t + 11t - 16t\right) + \left(-25s + 33s - 8s\right)\\[1mm] &= -25 + 0t + 0s\\[1mm] &= -25. \end{aligned} $

Taking the absolute value yields:

$ \left|5\alpha-11\beta-8\gamma\right| = 25. $

Conclusion

The value of

$ \left|5\,\alpha - 11\,\beta - 8\,\gamma\right| $ is $ \boxed{25}. $

$\begin{aligned} & L=\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} \\ & P Q=\frac{x-\frac{15}{7}}{1}=\frac{y-\frac{32}{7}}{4}=\frac{z-7}{7}=\lambda \end{aligned}$

$\begin{aligned} &\Rightarrow \mathrm{Q}\left(\lambda+\frac{15}{7}, 4 \lambda+\frac{32}{7}, 7 \lambda+7\right)\\ &\text { Since } \mathrm{Q} \text { lies on line } \mathrm{L}\\ &\begin{aligned} & \text { So, } \frac{\lambda+\frac{15}{7}+1}{3}=\frac{7 \lambda+7+5}{7} \\ & \Rightarrow 7 \lambda+22=21 \lambda+36 \\ & \Rightarrow \lambda=-1 \\ & \therefore \text { Point } Q\left(\frac{8}{7}, \frac{4}{7}, 0\right) \\ & \mathrm{PQ}=\sqrt{\left(\frac{15}{7}-\frac{8}{7}\right)^2+\left(\frac{32}{7}-\frac{4}{7}\right)^2+(7-0)} \\ & \mathrm{PQ}=\sqrt{66} \\ & \Rightarrow(\mathrm{PQ})^2=66 \end{aligned} \end{aligned}$

$\begin{aligned} & A(x, y, z) \text { Let } P(0,3,2), Q(2,0,3), R(0,0,1) \\ & A P=A Q=A R \\ & x^2+(y-3)^2+(z-2)^2=(x-2)^2+y^2+(z-3)^2=x^2+ \\ & y^2+(z-1)^2 \end{aligned}$

In xy plane $\mathrm{z}=0$

So, $x^2-4 x+4+y^2+9=x^2+y^2+1$

$\begin{aligned} & \Rightarrow y=2 \\ & x=3 \\ & 9+y^2-6 y+9+4=x^2+y^2+1 \end{aligned}$

So, $A(3,2,0)$ also $B(1,4,-1) \& C(2,0,-2)$

Now $A B=\sqrt{4+4+1}=3$

$\begin{aligned} &\begin{aligned} & \mathrm{AC}=\sqrt{1+4+4}=3 \\ & \mathrm{BC}=\sqrt{1+16+1}=\sqrt{18} \\ & \mathrm{AB}=\mathrm{AC} \\ & \text { isosceles } \Delta \& \mathrm{AB}^2+\mathrm{AC}^2=\mathrm{BC}^2 \\ & \text { right angle } \Delta \\ & \text { Area of } \triangle \mathrm{ABC}=\frac{1}{2} \times \text { base.height } \\ & \frac{1}{2} \times 3 \times 3=\frac{9}{2} \end{aligned}\\ &\text { So only } S_1 \text { is true } \end{aligned}$

$\begin{aligned} &\begin{aligned} & \overrightarrow{\mathrm{PQ}} \perp(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ & \Rightarrow 2(2 \lambda-3)+1(\lambda-2)+3(3 \lambda-2)=0 \\ & \Rightarrow 14 \lambda-14=0, \lambda=1 \\ & \text { So, } \mathrm{Q}(3,3,4) \end{aligned}\\ &\text { Let image in } R(\alpha, \beta, \gamma)\\ &\begin{aligned} & \frac{\alpha+4}{2}=3, \frac{\beta+4}{2}=3, \frac{\gamma+3}{2}=4 \\ & (\alpha, \beta, \gamma)=(2,2,5) \\ & \Rightarrow \alpha+\beta+\gamma=9 \end{aligned} \end{aligned}$

$\begin{aligned} & \text { Let } \mathrm{M}(3 \lambda+6,2 \lambda+7,-2 \lambda+7) \\ & \overrightarrow{\mathrm{BM}}=(3 \lambda+5) \hat{\mathrm{i}}+(2 \lambda+5) \hat{\mathrm{j}}+(-2 \lambda+4) \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BM}}=0=3(3 \lambda+5)+2(2 \lambda+5)-2(-2 \lambda+4) \\ & \overrightarrow{\mathrm{BM}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} \\ & \mid \overrightarrow{\mathrm{BM}}=7 \\ & \text { Area }=\frac{1}{2} \times 6 \times 7=21 \end{aligned}$

Equation of line through point $(-1,2,1)$ is $\rightarrow$

$\begin{aligned} &\Rightarrow \frac{x+1}{2}=\frac{y-2}{3}=\frac{z-1}{4}-(2)=\lambda\\ &\text { So, }\left[\begin{array}{l} x=2 \lambda-1 \\ y=3 \lambda+2 \\ z=4 \lambda+1 \end{array}\right. \end{aligned}$

By $(1) \rightarrow \frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}=\mu($ Let $)$

So, $\left[\begin{array}{l}x=3 \mu-2 \\ y=2 \mu+3 \\ z=\mu+4\end{array}\right.$

For intersection point 'P'

$\begin{aligned} & \mathrm{x}=2 \lambda-1=3 \mu-2 \\ & \mathrm{y}=3 \lambda+2=2 \mu+3 \quad\left[\begin{array}{l} \lambda=1 \\ \mu=1 \end{array}\right] \\ & \mathrm{z}=4 \lambda+1=\mu+4 \\ & \text { So, point } \mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(1,5,5) \\ & \& \mathrm{Q}(4,-5,1) \\ & \therefore \mathrm{PQ}=\sqrt{9+100+16} \\ & =\sqrt{125}=5 \sqrt{5} \end{aligned}$

$\begin{aligned} & \overrightarrow{\mathrm{a}}=(2,1,-3) \\ & \overrightarrow{\mathrm{b}}=(-1,-3,-5) \\ & \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{array}\right| \\ & =2 \hat{\mathrm{i}}-\hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{~b}}-\overrightarrow{\mathrm{a}}=-3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}} \\ & \mathrm{~S}_{\mathrm{d}}=\frac{|(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})|}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|} \\ & =\frac{2}{\sqrt{5}} \\ & \left(\mathrm{~S}_{\mathrm{d}}\right)^2=\frac{4}{5} \\ & \mathrm{~m}=4, \mathrm{n}=5 \Rightarrow \mathrm{~m}+\mathrm{n}=9 \end{aligned}$

To find the distance from the line

$ \frac{x-2}{2} = \frac{y-6}{3} = \frac{z-3}{4} $

to the point $(1,4,0)$ along the line

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

we first consider a parallel line passing through the point $(1,4,0)$. The equation of this parallel line is:

$ \frac{x-1}{1} = \frac{y-4}{2} = \frac{z-0}{3}. $

Next, we find the point of intersection between the given line and the parallel line, which can be expressed in parameter form as:

$ (\lambda + 1, 2\lambda + 4, 3\lambda) = (2t + 2, 3t + 6, 4t + 3). $

Solving for $\lambda$ in terms of $t$, from the first coordinate:

$ \lambda = 2t + 1. $

Using the second coordinates, we equate:

$ 2\lambda + 4 = 3t + 6 \quad \Rightarrow \quad 2(2t + 1) + 4 = 3t + 6. $

Simplifying gives:

$ 4t + 2 + 4 = 3t + 6 \quad \Rightarrow \quad 4t + 6 = 3t + 6 \quad \Rightarrow \quad t = 0. $

Substituting $t = 0$ into either line equation yields the point of intersection (POI) as:

$ (2, 6, 3). $

The distance from the point $(1, 4, 0)$ to the POI $(2, 6, 3)$ is computed as:

$ \sqrt{(2-1)^2 + (6-4)^2 + (3-0)^2} = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}. $

$\begin{aligned} &\begin{aligned} & \frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}=\lambda \\ & \Rightarrow 7 \lambda+3,-\lambda+2,-2 \lambda-1 \\ & \text { dr's of QP } \Rightarrow \\ & 7 \lambda-7,-\lambda+5,-2 \lambda \end{aligned}\\ &\text { Now }\\ &\begin{aligned} & (7 \lambda-7) \cdot 7-(-\lambda+5)+(2 \lambda) \cdot 2=0 \\ & 54 \lambda-54=0 \Rightarrow \lambda=1 \\ & \therefore \mathrm{P}=(10,1,-3) \\ & \overrightarrow{\mathrm{PQ}}=-4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{PR}}=-7 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \end{aligned} \end{aligned}$

Area $=\left| {{1 \over 2}\left| {\matrix{ i & j & k \cr 0 & { - 4} & 2 \cr { - 7} & { - 3} & 4 \cr } } \right|} \right| = 3\sqrt {30} $

$ P = (-2, -1, 3) $

Since the line through $ P $ is parallel to the vector

$ \vec{d} = (3, 2, 2), $

any point $ Q $ on this line can be expressed as:

$ Q = P + t\,\vec{d} = (-2 + 3t,\, -1 + 2t,\, 3 + 2t), $

where $ t $ is a real number. Given that $ P $ and $ Q $ must be distinct, we require $ t \neq 0 $.

The point $ R $ is given by:

$ R = (1, 3, 3). $

The distance from $ Q $ to $ R $ is $ 5 $, so we have:

$ \left[ (-2 + 3t - 1)^2 + (-1 + 2t - 3)^2 + (3 + 2t - 3)^2 \right] = 5^2. $

Simplify each coordinate difference:

For the $ x $-coordinate:

$ -2 + 3t - 1 = 3t - 3 = 3(t - 1). $

For the $ y $-coordinate:

$ -1 + 2t - 3 = 2t - 4 = 2(t - 2). $

For the $ z $-coordinate:

$ 3 + 2t - 3 = 2t. $

So the equation becomes:

$ [3(t - 1)]^2 + [2(t - 2)]^2 + (2t)^2 = 25. $

Expanding the squares:

$ 9(t - 1)^2 + 4(t - 2)^2 + 4t^2 = 25. $

Expand each term:

$ 9(t^2 - 2t + 1) + 4(t^2 - 4t + 4) + 4t^2 = 25, $

which simplifies to:

$ 9t^2 - 18t + 9 + 4t^2 - 16t + 16 + 4t^2 = 25. $

Combine like terms:

$ (9t^2 + 4t^2 + 4t^2) - (18t + 16t) + (9 + 16) = 25, $

$ 17t^2 - 34t + 25 = 25. $

Subtract 25 from both sides:

$ 17t^2 - 34t = 0. $

Factor out $ 17t $:

$ 17t(t - 2) = 0. $

Since $ t \neq 0 $, we must have:

$ t = 2. $

Substitute $ t = 2 $ back into the equation for $ Q $:

$ Q = (-2 + 3(2),\, -1 + 2(2),\, 3 + 2(2)) = (4, 3, 7). $

Next, to find the square of the area of triangle $ PQR $, first compute the vectors:

$ \vec{PQ} = Q - P = (4 - (-2),\, 3 - (-1),\, 7 - 3) = (6, 4, 4), $

$ \vec{PR} = R - P = (1 - (-2),\, 3 - (-1),\, 3 - 3) = (3, 4, 0). $

The area of the triangle is given by:

$ \text{Area} = \frac{1}{2} \, \| \vec{PQ} \times \vec{PR} \|. $

Calculate the cross product:

$ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 4 & 4 \\ 3 & 4 & 0 \\ \end{vmatrix}. $

Expanding the determinant:

$ \hat{i} $-component:

$ 4 \times 0 - 4 \times 4 = -16, $

$ \hat{j} $-component:

$ -(6 \times 0 - 4 \times 3) = 12, $

$ \hat{k} $-component:

$ 6 \times 4 - 4 \times 3 = 24 - 12 = 12. $

Thus,

$ \vec{PQ} \times \vec{PR} = (-16,\, 12,\, 12). $

Find the magnitude squared of the cross product:

$ \| \vec{PQ} \times \vec{PR} \|^2 = (-16)^2 + 12^2 + 12^2 = 256 + 144 + 144 = 544. $

The square of the area of triangle $ PQR $ is:

$ (\text{Area})^2 = \left(\frac{1}{2}\right)^2 \| \vec{PQ} \times \vec{PR} \|^2 = \frac{1}{4} \times 544 = 136. $

Thus, the square of the area of $ \triangle PQR $ is $136.$

To find the perpendicular distance from the point $P(2,-10,1)$ to the line given by

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

follow these steps:

Parametrize the Line:

From the given symmetric equations, set the common parameter as $t$:

$x = 1 + 2t$

$y = -2 - t$

$z = -3 + 2t$

This shows that:

A point on the line is $A(1,-2,-3)$ (when $t=0$).

The direction vector is $\vec{d} = \langle 2, -1, 2 \rangle.$

Determine the Vector from Point A to P:

Calculate

$\vec{AP} = P - A = \langle 2 - 1, \; -10 - (-2), \; 1 - (-3) \rangle = \langle 1, \; -8, \; 4 \rangle.$

Compute the Cross Product:

The formula for the perpendicular distance from a point to a line in 3D is:

$d = \frac{\|\vec{AP} \times \vec{d}\|}{\|\vec{d}\|}.$

First, find the cross product $\vec{AP} \times \vec{d}$:

$ \vec{AP} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -8 & 4 \\ 2 & -1 & 2 \end{vmatrix} = \Big( (-8)(2) - (4)(-1), \; (4)(2) - (1)(2), \; (1)(-1) - (-8)(2) \Big). $

Evaluate each component:

First component: $(-8 \times 2) - (4 \times -1) = -16 + 4 = -12.$

Second component: $ (4 \times 2) - (1 \times 2) = 8 - 2 = 6.$

Third component: $ (1 \times -1) - (-8 \times 2) = -1 + 16 = 15.$

So,

$\vec{AP} \times \vec{d} = \langle -12, 6, 15 \rangle.$

Calculate the Magnitudes:

For the cross product:

$ \|\vec{AP} \times \vec{d}\| = \sqrt{(-12)^2 + 6^2 + 15^2} = \sqrt{144 + 36 + 225} = \sqrt{405} = 9\sqrt{5}. $

For the direction vector:

$ \|\vec{d}\| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3. $

Compute the Distance:

Substitute the magnitudes into the distance formula:

$ d = \frac{9\sqrt{5}}{3} = 3\sqrt{5}. $

Thus, the perpendicular distance from the point $P(2,-10,1)$ to the line is $3\sqrt{5}$.

Comparing with the options given, the correct answer is:

Option C: $3\sqrt{5}$.

$\begin{aligned} & L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \\\\ & L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\end{aligned}$

$ \begin{aligned} & P(2 \lambda+1,3 \lambda+2,4 \lambda+3) \\ & Q(3 \mu+2,4 \mu+4,5 \mu+5) \end{aligned} $

Dr's of $P Q<2 \lambda-3 \mu-1,3 \lambda-4 \mu-2$,

$ \begin{aligned} & 4 \lambda-5 \mu-2> \\ & P Q=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}\right|=-\hat{i}+2 \hat{j}-\hat{k} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{2 \lambda-3 \mu-1}{-1}=\frac{3 \lambda-4 \mu-2}{2}=\frac{4 \lambda-5 \mu-2}{-1} \\ & \Rightarrow \quad \lambda=\frac{1}{3} \mu=\frac{-1}{6} \\ & \Rightarrow \quad P\left(\frac{5}{3}, 3, \frac{13}{3}\right) \quad Q\left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right) \end{aligned} $

Dr's $P Q\langle 1,-2,1\rangle$

$\therefore$ Line

$ \frac{y-\frac{5}{3}}{1}=\frac{y-3}{-2}=\frac{y-\frac{13}{3}}{1} $

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} $

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

Measured along $L_2: \frac{x-\frac{11}{3}}{\frac{2}{3}}=\frac{y-\frac{11}{3}}{\frac{1}{3}}=\frac{z-\frac{19}{3}}{\frac{2}{3}}=\lambda$

Any point on $L_1:(\mu+1, \mu+2,2 \mu+3)$

Any point on $L_2\left(\frac{2}{3} \lambda+\frac{11}{3}, \frac{\lambda}{3}+\frac{11}{3}, \frac{2}{3} \lambda+\frac{19}{3}\right)$

Now

$\begin{aligned} & \mu+1=\frac{2}{3} \lambda+\frac{11}{3} \\ & \frac{\mu+2=\frac{\lambda}{3}+\frac{11}{3}}{\lambda=-3} \\ & \mu=\frac{2}{3} \\ \end{aligned}$

$\begin{aligned} & \text { Point on } L=\left(\frac{5}{3}, \frac{8}{3}, \frac{13}{3}\right) \\ & d=\sqrt{\left(\frac{11}{3}-\frac{5}{3}\right)^2+\left(\frac{8}{3}-\frac{11}{3}\right)^2+\left(\frac{19}{3}-\frac{13}{3}\right)^2} \\ & d=\sqrt{4+1+4} \\ & d=3 \end{aligned}$