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} $

$\frac{x-4}{1}=\frac{y-9}{2}=\frac{z-5}{1}$

$\mathrm{Q}(1, \mu, 2)$ satisfies the line L

$ \therefore \frac{1-4}{1}=\frac{\mu-9}{2}=\frac{2-5}{1} $

$ \Rightarrow-3=\frac{\mu-9}{2}=-3 \quad \Rightarrow \mu=3 $

D.R. of PQ : $\lambda-1,2-\mu, 1$

Since $\mathrm{PQ} \perp \mathrm{L} \Rightarrow(\lambda-1) .1+2(2-\mu)+1.1=0$

$ \Rightarrow \lambda-2 \mu+4=0 \quad \therefore \lambda=2 $

Now lines : $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$ \&

$ \frac{x-2}{2}=\frac{y-3}{3}=\frac{z+5}{6} $

$\begin{aligned} & \text { i.e. } \vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+t_1(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\ & \& \vec{r}=(2 \hat{i}+3 \hat{j}-5 \hat{k})+t_2(2 \hat{i}+3 \hat{j}+6 \hat{k})\end{aligned}$

Distance

$ \begin{aligned} & =\frac{|((\hat{i}+2 \hat{j}-4 \hat{k})-(2 \hat{i}+3 \hat{j}-5 \hat{k})) \times(2 \hat{i}+3 \hat{j}+6 \hat{k})|}{|2 \hat{i}+3 \hat{j}+6 \hat{k}|} \\ & =\frac{|-9 \hat{i}+8 \hat{j}-\hat{k}|}{7}=\frac{\sqrt{146}}{7} \end{aligned} $

For the two lines

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

and

$ \frac{x+2}{2}=\frac{y-6}{4}=\frac{z-5}{-5}, $

we use the NCERT formula for the shortest distance between two skew lines:

$ \text{Shortest distance}= \frac{\left|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times \vec{b_2})\right|}{\left|\vec{b_1}\times \vec{b_2}\right|} $

where:

• $\vec{a_1}$ is a point on the first line

• $\vec{a_2}$ is a point on the second line

• $\vec{b_1}, \vec{b_2}$ are direction vectors of the two lines

Step 1: Identify points and direction vectors

From the first line,

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

a point on it is

$ A(4,3,2) $

and its direction vector is

$ \vec{b_1}=\hat{i}+2\hat{j}-3\hat{k} $

From the second line,

$ \frac{x+2}{2}=\frac{y-6}{4}=\frac{z-5}{-5} $

a point on it is

$ B(-2,6,5) $

and its direction vector is

$ \vec{b_2}=2\hat{i}+4\hat{j}-5\hat{k} $

Step 2: Find $\vec{AB}$

$ \vec{AB}=B-A=(-2-4,\ 6-3,\ 5-2)=(-6,3,3) $

So,

$ \vec{AB}=-6\hat{i}+3\hat{j}+3\hat{k} $

Step 3: Find $\vec{b_1}\times \vec{b_2}$

$ \vec{b_1}\times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & 2 & -3\\ 2 & 4 & -5 \end{vmatrix} $

Expanding,

$ = \hat{i}\left(2(-5)-(-3)(4)\right) -\hat{j}\left(1(-5)-(-3)(2)\right) +\hat{k}\left(1\cdot 4-2\cdot 2\right) $

$ = \hat{i}(-10+12)-\hat{j}(-5+6)+\hat{k}(4-4) $

$ = 2\hat{i}-\hat{j}+0\hat{k} $

So,

$ \vec{b_1}\times \vec{b_2}=(2,-1,0) $

Its magnitude is

$ \left|\vec{b_1}\times \vec{b_2}\right| =\sqrt{2^2+(-1)^2+0^2} =\sqrt{5} $

Step 4: Find scalar triple product

$ \vec{AB}\cdot(\vec{b_1}\times \vec{b_2}) =(-6,3,3)\cdot(2,-1,0) $

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

$ = -12-3=-15 $

So,

$ \left| \vec{AB}\cdot(\vec{b_1}\times \vec{b_2}) \right|=15 $

Step 5: Find shortest distance

$ d=\frac{15}{\sqrt{5}}=3\sqrt{5} $

Hence, the shortest distance is

$ \boxed{3\sqrt{5}} $

So, the correct option is:

$ \boxed{\text{Option C}} $

Let the line be written in parametric form as

$ L:\ \frac{x}{1}=\frac{y-1}{2}=\frac{z-a+1}{b}=t $

So a general point on $L$ is

$ Q(t)=(t,\ 1+2t,\ a-1+bt). $

We are given:

• Point $P=(1,6,a)$

• Its image in the line $L$ is $\left(\frac a3,0,a+c\right)$

For reflection in a line in 3D, the line is the perpendicular bisector of the segment joining the point and its image.

So if image of $P$ is $P'=\left(\frac a3,0,a+c\right)$, then:

1. Midpoint of $PP'$ lies on $L$

2. The line $L$ is perpendicular to $\overrightarrow{PP'}$

1. Midpoint of $P$ and $P'$

Midpoint $M$ is

$ M=\left(\frac{1+\frac a3}{2},\frac{6+0}{2},\frac{a+(a+c)}{2}\right) =\left(\frac{a+3}{6},3,\frac{2a+c}{2}\right) $

Since $M$ lies on $L$, for some parameter $t$,

$ (t,1+2t,a-1+bt)=\left(\frac{a+3}{6},3,\frac{2a+c}{2}\right) $

From the $y$-coordinate,

$ 1+2t=3 \implies 2t=2 \implies t=1 $

Then from the $x$-coordinate,

$ t=\frac{a+3}{6} $

So

$ 1=\frac{a+3}{6}\implies a+3=6\implies a=3 $

Now from the $z$-coordinate,

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

Substitute $a=3$:

$ 3-1+b=\frac{6+c}{2} $

$ 2+b=\frac{6+c}{2} $

$ 4+2b=6+c $

$ c=2b-2 $

2. Perpendicularity condition

Direction vector of line $L$ is

$ \vec d=(1,2,b) $

Now

$ \overrightarrow{PP'}=\left(\frac a3-1,\ 0-6,\ a+c-a\right) =\left(\frac a3-1,-6,c\right) $

Since $a=3$,

$ \overrightarrow{PP'}=(0,-6,c) $

This must be perpendicular to $\vec d=(1,2,b)$, so

$ (0,-6,c)\cdot(1,2,b)=0 $

$ 0-12+bc=0 $

$ bc=12 $

But we already have

$ c=2b-2 $

So

$ b(2b-2)=12 $

$ 2b^2-2b-12=0 $

$ b^2-b-6=0 $

$ (b-3)(b+2)=0 $

Given $b>0$, so

$ b=3 $

Then

$ c=2b-2=6-2=4 $

Thus the line is

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

and the image point is

$ P'=\left(1,0,7\right). $

3. Foot of perpendicular from $P$ to $L$

Let the foot be $F$. Since midpoint of $P$ and its image lies on the mirror line, $F$ is exactly the midpoint of $P$ and $P'$:

$ F=\left(\frac{1+1}{2},\frac{6+0}{2},\frac{3+7}{2}\right)=(1,3,5) $

Check that this lies on $L$:

For $t=1$,

$ Q(1)=(1,3,5) $

Correct.

4. Point $S(\alpha,\beta,\gamma)$ on $L$ such that $SF=2\sqrt{14}$

A general point on $L$ is

$ S(t)=(t,1+2t,2+3t) $

Since $F=(1,3,5)$ corresponds to $t=1$, we get

$ \overrightarrow{FS}=(t-1,\ 2t-2,\ 3t-3)=(t-1)(1,2,3) $

Hence

$ SF=\sqrt{(t-1)^2+(2t-2)^2+(3t-3)^2} $

$ =|t-1|\sqrt{1+4+9}=|t-1|\sqrt{14} $

Given

$ SF=2\sqrt{14} $

So

$ |t-1|=2 $

Thus

$ t=3 \quad \text{or} \quad t=-1 $

Since $\alpha>0$ and $\alpha=x=t$, we need

$ t>0 \implies t=3 $

Therefore

$ S=(3,7,11) $

So

$ \alpha+\beta+\gamma=3+7+11=21 $

Hence the correct answer is

$ \boxed{21} $

So, Option C is correct.

For a line $L$ to be perpendicular to both lines $L_1$ and $L_2$, its direction ratios must be perpendicular to the direction ratios of both $L_1$ and $L_2$.

So first let us find the direction vectors.

For $L_1$:

$ \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} $

Direction ratios of $L_1$ are

$ \vec{d_1}=(3,5,7) $

For $L_2$:

$ \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7} $

Direction ratios of $L_2$ are

$ \vec{d_2}=(1,4,7) $

Since line $L$ is perpendicular to both, its direction vector will be parallel to

$ \vec{d}=\vec{d_1}\times \vec{d_2} $

Now,

$ \vec{d_1}\times \vec{d_2} = \begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & 5 & 7\\ 1 & 4 & 7 \end{vmatrix} $

$ = \hat i(35-28)-\hat j(21-7)+\hat k(12-5) $

$ = 7\hat i-14\hat j+7\hat k $

$ =7(1,-2,1) $

So direction ratios of line $L$ can be taken as

$ (1,-2,1) $

Now for $L_3$:

$ \frac{x-\frac{8}{7}}{2}=\frac{y-\frac{4}{7}}{1}=\frac{z}{2} $

Direction ratios of $L_3$ are

$ (2,1,2) $

Let $\theta$ be the acute angle between $L$ and $L_3$.

Using the formula

$ \tan\theta=\frac{|\vec a\times \vec b|}{|\vec a\cdot \vec b|} $

where

$ \vec a=(1,-2,1), \qquad \vec b=(2,1,2) $

First find dot product:

$ \vec a\cdot \vec b = 1\cdot 2+(-2)\cdot 1+1\cdot 2 $

$ =2-2+2=2 $

Now find cross product:

$ \vec a\times \vec b= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -2 & 1\\ 2 & 1 & 2 \end{vmatrix} $

$ =\hat i((-2)(2)-1\cdot 1)-\hat j(1\cdot 2-1\cdot 2)+\hat k(1\cdot 1-(-2)\cdot 2) $

$ =\hat i(-4-1)-\hat j(2-2)+\hat k(1+4) $

$ =-5\hat i+0\hat j+5\hat k $

So,

$ |\vec a\times \vec b| = \sqrt{(-5)^2+0^2+5^2} = \sqrt{50} = 5\sqrt{2} $

Hence,

$ \tan\theta=\frac{5\sqrt{2}}{2} $

Therefore,

$ \boxed{\tan\theta=\frac{5}{2}\sqrt{2}} $

So the correct option is:

$ \boxed{\text{Option B}} $

Write the given line in parametric form.

$ \frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}=t $

So any point on the line is

$ P(t) = (-3+8t,\; 4+2t,\; -1+2t) $

Since both $P$ and $Q$ lie on this line and are at a distance $6$ from $R(1,2,3)$, we find the points on the line whose distance from $R$ is $6$.

Distance condition:

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

Substitute $x=-3+8t,\; y=4+2t,\; z=-1+2t$:

$ (-3+8t-1)^2+(4+2t-2)^2+(-1+2t-3)^2=36 $

$ (8t-4)^2+(2t+2)^2+(2t-4)^2=36 $

Now expand:

$ 64t^2-64t+16 + 4t^2+8t+4 + 4t^2-16t+16 = 36 $

$ 72t^2-72t+36=36 $

$ 72t^2-72t=0 $

$ 72t(t-1)=0 $

So,

$ t=0 \quad \text{or} \quad t=1 $

Hence the two points are:

For $t=0$,

$ P=(-3,4,-1) $

For $t=1$,

$ Q=(5,6,1) $

Now centroid of triangle $PQR$ is

$ (\alpha,\beta,\gamma)=\left(\frac{x_P+x_Q+x_R}{3},\frac{y_P+y_Q+y_R}{3},\frac{z_P+z_Q+z_R}{3}\right) $

Substitute $P=(-3,4,-1)$, $Q=(5,6,1)$, $R=(1,2,3)$:

$ \alpha=\frac{-3+5+1}{3}=\frac{3}{3}=1 $

$ \beta=\frac{4+6+2}{3}=\frac{12}{3}=4 $

$ \gamma=\frac{-1+1+3}{3}=\frac{3}{3}=1 $

Therefore,

$ \alpha+\beta+\gamma = 1+4+1 = 6 $

So the correct answer is:

$ \boxed{6} $

Hence, Option C is correct.

Let the point $P(1,2,7)$.

We need the image of $P$ in the line

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

This means the line passes through the point

$ A(0,1,2) $

and has direction ratios

$ \vec d=(1,1,2). $

If $P'$ is the image of $P$ in this line, then the given line is the perpendicular bisector of the segment joining $P$ and $P'$ in the sense of reflection about a line in 3D. So first we find the foot of perpendicular from $P$ to the line.

Let the foot be $M$.

A general point on the line is

$ M(t)=(t,1+t,2+2t). $

Now,

$ \overrightarrow{PM}=(t-1,\; t-1,\; 2t-5). $

Since $PM$ is perpendicular to the line, we must have

$ \overrightarrow{PM}\cdot \vec d=0 $

So,

$ (t-1)(1)+(t-1)(1)+(2t-5)(2)=0 $

$ t-1+t-1+4t-10=0 $

$ 6t-12=0 $

$ t=2 $

Hence,

$ M=(2,3,6). $

Since $M$ is the midpoint of $P(1,2,7)$ and its image $P'$, we get

$ P'=2M-P $

$ P'=(4,6,12)-(1,2,7)=(3,4,5). $

So the image of $(1,2,7)$ in the given line is

$ (3,4,5). $

Now the distance of $(a,2,5)$ from this image is $4$:

$ \sqrt{(a-3)^2+(2-4)^2+(5-5)^2}=4 $

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

Squaring,

$ (a-3)^2+4=16 $

$ (a-3)^2=12 $

$ a-3=\pm 2\sqrt{3} $

So,

$ a=3\pm 2\sqrt{3}. $

Therefore, the sum of all possible values of $a$ is

$ (3+2\sqrt{3})+(3-2\sqrt{3})=6. $

Hence, the correct option is

$ \boxed{6} $

So, option C is correct.

Write the given line in vector form.

From

$ \frac{x-2}{2}=\frac{y-5}{3}=\frac{z-2}{4}, $

a point on the line is

$ A(2,5,2) $

and its direction vector is

$ \vec{b} = 2\hat{i}+3\hat{j}+4\hat{k}. $

Given point is

$ P(5,6,7). $

Now,

$ \overrightarrow{AP} = (5-2)\hat{i}+(6-5)\hat{j}+(7-2)\hat{k} = 3\hat{i}+\hat{j}+5\hat{k}. $

The distance $d$ of point $P$ from the line is given by

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

Since we need the square of the distance,

$ d^2=\frac{|\overrightarrow{AP}\times \vec{b}|^2}{|\vec{b}|^2}. $

Now calculate the cross product:

$ \overrightarrow{AP}\times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 1 & 5\\ 2 & 3 & 4 \end{vmatrix} $

$ = \hat{i}(1\cdot 4-5\cdot 3)-\hat{j}(3\cdot 4-5\cdot 2)+\hat{k}(3\cdot 3-1\cdot 2) $

$ = \hat{i}(4-15)-\hat{j}(12-10)+\hat{k}(9-2) $

$ = -11\hat{i}-2\hat{j}+7\hat{k}. $

So,

$ |\overrightarrow{AP}\times \vec{b}|^2 = (-11)^2+(-2)^2+7^2 = 121+4+49=174. $

Also,

$ |\vec{b}|^2 = 2^2+3^2+4^2=4+9+16=29. $

Therefore,

$ d^2=\frac{174}{29}=6. $

So, the square of the distance is

$ \boxed{6} $

Hence, the correct option is Option C.

For the first line, the equation is:

$ \vec{r}_1 = (\hat{i} + \hat{j} - \hat{k}) + \lambda(a \hat{i} - \hat{j}) $

We can group the components along the $\hat{i}$, $\hat{j}$, and $\hat{k}$ directions:

$ \vec{r}_1 = (1 + a\lambda)\hat{i} + (1 - \lambda)\hat{j} - \hat{k} $

For the second line, the equation is:

$ \vec{r}_2 = (4 \hat{i} - \hat{k}) + \mu(2 \hat{i} + a \hat{k}) $

Grouping the components similarly:

$ \vec{r}_2 = (4 + 2\mu)\hat{i} + 0\hat{j} + (a\mu - 1)\hat{k} $

If the two lines intersect, there must be a common point. This means that for some value of $\lambda$ and $\mu$, their position vectors will be exactly the same. We can find this by equating the corresponding components of $\vec{r}_1$ and $\vec{r}_2$:

1. Equating the $\hat{j}$ components:

$ 1 - \lambda = 0 $

$ \lambda = 1 $

2. Equating the $\hat{k}$ components:

$ -1 = a\mu - 1 $

$ a\mu = 0 $

Since the question specifies that $a \neq 0$, it must be true that $\mu = 0$.

3. Equating the $\hat{i}$ components:

$ 1 + a\lambda = 4 + 2\mu $

Now, substitute the values of $\lambda = 1$ and $\mu = 0$ into this equation:

$ 1 + a(1) = 4 + 2(0) $

$ 1 + a = 4 $

$ a = 3 $

Now that we have $\mu = 0$ (and $\lambda = 1$), we can substitute either of these back into the respective line equations to find the position vector of the point of intersection. Let us use the second line's equation by substituting $\mu = 0$:

$ \vec{r} = (4 + 2(0))\hat{i} + (3(0) - 1)\hat{k} $

$ \vec{r} = 4\hat{i} - \hat{k} $

So, the coordinates of the point of intersection are $(4, 0, -1)$.

Next, we calculate the distance $d$ of this point of intersection from the origin $(0,0,0)$ using the distance formula:

$ d = \sqrt{x^2 + y^2 + z^2} $

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

$ d = \sqrt{16 + 0 + 1} $

$ d = \sqrt{17} $

Looking at the given options, the problem asks for the square of the distance from the origin.

The square of the distance is:

$ d^2 = (\sqrt{17})^2 = 17 $

Therefore, the correct choice is Option C.

For two skew lines,

$ \vec r=\vec a_1+\lambda \vec b_1 \quad \text{and} \quad \vec r=\vec a_2+\mu \vec b_2, $

the shortest distance is

$ d=\frac{|(\vec a_2-\vec a_1)\cdot (\vec b_1\times \vec b_2)|}{|\vec b_1\times \vec b_2|}. $

Now compare with the given lines.

First line:

$ \vec r=\left(\frac13 \hat i+2\hat j+\frac83 \hat k\right)+\lambda(2\hat i-5\hat j+6\hat k) $

So,

$ \vec a_1=\left(\frac13,2,\frac83\right), \qquad \vec b_1=(2,-5,6) $

Second line:

$ \vec r=\left(-\frac23 \hat i-\frac13 \hat k\right)+\mu(\hat j-\hat k) $

So,

$ \vec a_2=\left(-\frac23,0,-\frac13\right), \qquad \vec b_2=(0,1,-1) $

Now find

$ \vec a_2-\vec a_1 =\left(-\frac23-\frac13,\;0-2,\;-\frac13-\frac83\right) =(-1,-2,-3) $

Next, calculate $\vec b_1\times \vec b_2$:

$ \vec b_1\times \vec b_2 = \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & -5 & 6\\ 0 & 1 & -1 \end{vmatrix} $

$ = \hat i\big[(-5)(-1)-6(1)\big] -\hat j\big[2(-1)-6(0)\big] +\hat k\big[2(1)-(-5)(0)\big] $

$ = \hat i(5-6)-\hat j(-2)+\hat k(2) = -\hat i+2\hat j+2\hat k $

So,

$ \vec b_1\times \vec b_2=(-1,2,2) $

Its magnitude is

$ |\vec b_1\times \vec b_2| =\sqrt{(-1)^2+2^2+2^2} =\sqrt{1+4+4} =3 $

Now find the scalar triple product:

$ (\vec a_2-\vec a_1)\cdot (\vec b_1\times \vec b_2) =(-1,-2,-3)\cdot(-1,2,2) $

$ = (-1)(-1)+(-2)(2)+(-3)(2) =1-4-6 =-9 $

Therefore,

$ d=\frac{|-9|}{3}=3 $

Hence, the shortest distance is

$ \boxed{3} $

So, the correct option is Option B.

$\therefore \mathrm{P}$ is mid point of AB and lies on the given line

$ \Rightarrow \mathrm{P}\left(\frac{3 \alpha+1}{2}, \frac{\alpha^2-\alpha}{2}, \frac{\alpha+1}{4}\right) $

Now put coordinates of P in the line

$ \begin{aligned} & \Rightarrow \frac{\frac{3 \alpha+1}{2}-2}{3}=\frac{\frac{\alpha^2-\alpha}{2}-1}{2}=\frac{\alpha+1}{4} \\ & \Rightarrow \frac{\alpha-1}{2}=\frac{\alpha+1}{4}=\frac{\alpha^2-\alpha-2}{4} \\ & \Rightarrow \alpha=3 \end{aligned} $

Now for $\alpha=3$, clearly AB is perpendicular to $3 \hat{i}+2 \hat{j}+\mathrm{k}$

for $\alpha=3$ point B is image of point A

P$(2\lambda, 3\lambda, 3\lambda - 1)$ and Q$(-\mu - 1, \mu + 2, 4\mu)$

D.R of PQ $2\lambda + \mu + 1$, $3\lambda - \mu - 2$, $3\lambda - 4\mu - 1$,

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

Solving $\lambda = \frac{1}{5}$ and $\mu = -\frac{8}{15}$

P$\left(\frac{2}{5}, \frac{3}{5}, -\frac{2}{5}\right)$ & Q$\left(-\frac{7}{15}, \frac{22}{15}, -\frac{32}{15}\right)$

$\mathrm{PQ}^2 = \left(\frac{13}{15}\right)^2 + \left(\frac{13}{15}\right)^2 + \left(\frac{26}{15}\right)^2$

$\alpha^2 = \frac{1014}{225} \therefore 225\alpha^2 = 1014$

Let the given point be $P(-2, -8, 6)$. We are looking for the distance of point $P$ from the first line, measured parallel to the second line.

Let the first line be $L_1$:

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

And the second line be $L_2$:

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

The direction ratios of the line $L_2$ are $(1, -1, 2)$. Since we need to measure the distance parallel to $L_2$, we will form the equation of a new line, say $L_3$, which passes through the point $P(-2, -8, 6)$ and is parallel to $L_2$.

The direction ratios of $L_3$ will be the same as those of $L_2$. Using the point-direction form from 3D geometry, the equation of the line $L_3$ is:

$ \frac{x - (-2)}{1} = \frac{y - (-8)}{-1} = \frac{z - 6}{2} $

$ \frac{x + 2}{1} = \frac{y + 8}{-1} = \frac{z - 6}{2} $

Let the line $L_3$ intersect the line $L_1$ at a point $Q$. To find the coordinates of $Q$, we can take a general point on $L_1$.

Let:

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

From this, the coordinates of any general point on $L_1$ can be expressed in terms of $\lambda$:

$ x = \lambda + 1 $

$ y = 2\lambda + 1 $

$ z = -\lambda $

So, we can assume the intersection point is $Q(\lambda + 1, 2\lambda + 1, -\lambda)$.

Since $Q$ also lies on the line $L_3$, its coordinates must satisfy the equation of $L_3$. Substituting the coordinates of $Q$ into the equation of $L_3$, we get:

$ \frac{(\lambda + 1) + 2}{1} = \frac{(2\lambda + 1) + 8}{-1} = \frac{-\lambda - 6}{2} $

$ \frac{\lambda + 3}{1} = \frac{2\lambda + 9}{-1} = \frac{-\lambda - 6}{2} $

Now, let us solve for $\lambda$ by equating the first two expressions:

$ \lambda + 3 = -(2\lambda + 9) $

$ \lambda + 3 = -2\lambda - 9 $

$ 3\lambda = -12 $

$ \lambda = -4 $

We can verify this value by substituting $\lambda = -4$ into the third expression:

$ \frac{-(-4) - 6}{2} = \frac{4 - 6}{2} = \frac{-2}{2} = -1 $

This matches the first part $ (-4) + 3 = -1 $. Thus, $\lambda = -4$ is correct.

Now, substitute $\lambda = -4$ back into our assumption to find the exact coordinates of $Q$:

$ x = (-4) + 1 = -3 $

$ y = 2(-4) + 1 = -7 $

$ z = -(-4) = 4 $

So, the point of intersection is $Q(-3, -7, 4)$.

We are asked to find the square of the distance between point $P$ and the line $L_1$ measured along $L_3$. This naturally corresponds to the square of the distance between points $P(-2, -8, 6)$ and $Q(-3, -7, 4)$.

Using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$:

$ PQ^2 = (-3 - (-2))^2 + (-7 - (-8))^2 + (4 - 6)^2 $

$ PQ^2 = (-3 + 2)^2 + (-7 + 8)^2 + (-2)^2 $

$ PQ^2 = (-1)^2 + (1)^2 + (-2)^2 $

$ PQ^2 = 1 + 1 + 4 $

$ PQ^2 = 6 $

The square of the required distance is $6$.

Option B

Let the given line be written in parametric form by taking

$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-\alpha}{3}=r.$

Then a general point $A$ on the line is

$A(1+2r,\;2+r,\;\alpha+3r).$

Given $P(a,b,0)$ and the midpoint of $PA$ is $\left(0,\frac{3}{4},-\frac{1}{4}\right)$. Using midpoint formula coordinate-wise:

$\begin{aligned} & \frac{2 r+1+a}{2}=0 \Rightarrow 2 r+a=-1 \\\\ & \frac{r+2+b}{2}=\frac{3}{4} \Rightarrow r+b=-\frac{1}{2}\end{aligned}$

Similarly, from the $z$-coordinate of the midpoint:

$\begin{aligned} & \frac{3 r+\alpha}{2}=\frac{-1}{4} \Rightarrow 3 r+\alpha=\frac{-1}{2} \\\\ \end{aligned}$

Now, $A$ is the foot of the perpendicular from $P$ to the line, so $\overrightarrow{PA}$ is perpendicular to the direction vector of the line.

The direction ratios of the line are $(2,1,3)$, so we use the dot product condition:

$2(x_A-a)+1(y_A-b)+3(z_A-0)=0.$

Using the midpoint point $\left(0,\frac{3}{4},-\frac{1}{4}\right)$, we can write $\overrightarrow{PA}=2\left(\left(0,\frac{3}{4},-\frac{1}{4}\right)-P\right)$, so its components are proportional to $\left(-a,\frac{3}{4}-b,-\frac{1}{4}\right)$. Hence the perpendicularity condition becomes:

$\begin{aligned} & 2 \cdot a+1 \cdot(b-3 / 4)+3 \cdot \frac{1}{4}=0 \Rightarrow a+b=0 \\\\ \end{aligned}$

From $2r+a=-1$, we get $a=-1-2r$.

From $r+b=-\frac{1}{2}$, we get $b=-\frac{1}{2}-r$.

Substitute in $a+b=0$:

$\begin{aligned} & \Rightarrow 2(-1-2 r)+(-r-1 / 2)=0 \\\\ & \Rightarrow-3 / 2=3 r=1 \Rightarrow r=-1 / 2, a=0, b=0, \alpha=1 \\\\ & \therefore a^2+b^2+\alpha^2=1\end{aligned}$

Let the common point of intersection of the two lines be $P$.

Since the point lies on the $xy$-plane, its $z$-coordinate must be $0$.

We write both lines in parametric form.

For the first line,

$ \frac{x+1}{3}=\frac{y+a}{5}=\frac{z+b+1}{7}=\lambda $

So,

$ x=3\lambda-1,\qquad y=5\lambda-a,\qquad z=7\lambda-b-1 $

For the second line,

$ \frac{x-2}{1}=\frac{y-b}{4}=\frac{z-2a}{7}=\mu $

So,

$ x=\mu+2,\qquad y=4\mu+b,\qquad z=7\mu+2a $

At the point of intersection, coordinates are equal.

So we get:

$ 3\lambda-1=\mu+2 $

$ 5\lambda-a=4\mu+b $

$ 7\lambda-b-1=7\mu+2a $

Also, since the point lies on the $xy$-plane,

$ z=0 $

Hence from both lines:

$ 7\lambda-b-1=0 $

and

$ 7\mu+2a=0 $

Now solve step by step.

From

$ 3\lambda-1=\mu+2 $

we get

$ 3\lambda-\mu=3 \qquad \Rightarrow \qquad \mu=3\lambda-3 $

From $z=0$ in the first line,

$ b=7\lambda-1 $

From $z=0$ in the second line,

$ 2a=-7\mu \qquad \Rightarrow \qquad a=-\frac{7\mu}{2} $

Now use

$ 5\lambda-a=4\mu+b $

Substitute $\mu=3\lambda-3$, $b=7\lambda-1$:

$ 5\lambda-a=4(3\lambda-3)+(7\lambda-1) $

$ 5\lambda-a=12\lambda-12+7\lambda-1 $

$ 5\lambda-a=19\lambda-13 $

$ -a=14\lambda-13 $

$ a=13-14\lambda $

Also,

$ a=-\frac{7\mu}{2}=-\frac{7(3\lambda-3)}{2} $

$ a=\frac{-21\lambda+21}{2} $

Equate the two values of $a$:

$ 13-14\lambda=\frac{-21\lambda+21}{2} $

Multiply by $2$:

$ 26-28\lambda=-21\lambda+21 $

$ 5=7\lambda $

$ \lambda=\frac{5}{7} $

Then

$ \mu=3\lambda-3=3\cdot \frac{5}{7}-3=\frac{15}{7}-\frac{21}{7}=-\frac{6}{7} $

Now find $a$ and $b$.

$ b=7\lambda-1=5-1=4 $

$ a=-\frac{7\mu}{2}=-\frac{7}{2}\left(-\frac{6}{7}\right)=3 $

Therefore,

$ a+b=3+4=7 $

So the correct answer is:

$ \boxed{7} $

Option C

Direction vector of line is

$ =\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{array}\right|=-2 \hat{i}+3 \hat{j}-2 \hat{k} $

So equation of line is

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

let ( $2 \lambda+1,-3 \lambda+1,2 \lambda+1$ ) be the foot of perpendicular from $(0,0,0)$ upon given line

$ \begin{aligned} & \Rightarrow \overrightarrow{\mathrm{OM}} \cdot(2 \hat{\mathrm{i}}-3 \mathrm{j}+2 \hat{\mathrm{k}})=0 \\ & \Rightarrow 2(2 \lambda+1)-3(-3 \lambda+1)+2(2 \lambda+1)=0 \\ & \Rightarrow \lambda=-\frac{1}{17} \end{aligned} $

So $(\mathrm{a}, \mathrm{b}, \mathrm{c}) \equiv\left(\frac{-2}{17}+1, \frac{3}{17}+1, \frac{-2}{17}+1\right)$

$ \Rightarrow 34(a+b+c)=100 $

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

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}. $

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

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