3D Geometry

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

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

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

Given lines are

$\frac{x-3}{4}=\frac{y-(-7)}{-11}=\frac{z-1}{5}$ and

$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z-(-2)}{1}$

Shortest distance between two lines,

$\begin{aligned} & d=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|} \\ & \vec{a}_2-\vec{a}_1=2 \hat{i}+16 \hat{j}-3 \hat{k} \text { and } \\ & \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & -11 & 5 \\ 3 & -6 & 1 \end{array}\right|=19 \hat{i}+11 \hat{j}+9 \hat{k} \\ & \therefore \quad d=\frac{187}{\sqrt{563}} \end{aligned}$

$\begin{aligned} & L_1: \frac{x-2}{1}=\frac{y}{-1}=\frac{z-1}{1}=\lambda \\ & L_2: \frac{x+1}{(1 / 2)}=\frac{y-1}{(1 / 2)}=\frac{z+1}{1}=\mu \end{aligned}$

Any point of $L_1$ and $L_2$ will be $(\lambda+2,-\lambda, \lambda+1)$ and $\left(\frac{\mu}{2}-1, \frac{\mu}{2}+1, \mu-1\right)$

Now Dr of line $<\lambda-\frac{\mu}{2}+3,-\lambda-\frac{\mu}{2}-1, \lambda-\mu+2>$

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

$\left. {\matrix{ {\lambda - {\mu \over 3} + 3 = - 3\lambda - {{3\mu } \over 2} - 3\,\,\,...(1)} \cr {2\left( {\lambda - {\mu \over 3} + 3} \right) = 3(\lambda - \mu + 2)\,\,...(2)} \cr } } \right\}\lambda = {{ - 4} \over 3},\mu = {{ - 2} \over 3}$

$\therefore \quad$ Points will be $\left(\frac{2}{3}, \frac{4}{3}, \frac{-1}{3}\right)$ and $\left(\frac{-4}{3}, \frac{2}{3}, \frac{-5}{3}\right)$

$\therefore \quad L$ will be $\frac{x-\frac{2}{3}}{3}=\frac{y-\frac{4}{3}}{1}=\frac{z+\frac{1}{3}}{2}$

$\therefore \quad\left(\frac{-1}{3}, 1,-1\right)$ will satisfy $L$

$\begin{aligned} & \vec{l}_1=2 \hat{i}+3 \hat{j}+4 \hat{k} \\ & \vec{l}_2=4 \hat{i}+6 \hat{j}+8 \hat{k} \end{aligned}$

$S . D .=\frac{|(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times((\lambda-2) \hat{i}-4 \hat{k})|}{|2 \hat{i}+3 \hat{j}+4 \hat{k}|}$

$\begin{aligned} & \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ \lambda-2 & 0 & -4 \end{array}\right|=-12 \hat{i}-\hat{j}(-8-4 \lambda+8)+\hat{k}(6-3 \lambda) \\ & =-12 \hat{i}+4 \lambda \hat{j}+(6-3 \lambda) \hat{k} \\ & \frac{\sqrt{144+16 \lambda^2+(6-3 \lambda)^2}}{\sqrt{29}}=\frac{13}{\sqrt{29}} \\ & 144+16 \lambda^2+(6-3 \lambda)^2=169 \\ & \Rightarrow 16 \lambda^2+9 \lambda^2+36-36 \lambda+144-169=0 \\ \end{aligned}$

$\begin{aligned} & \Rightarrow 25 \lambda^2-36 \lambda+11=0 \\ & \Rightarrow 25 \lambda^2-25 \lambda-11 \lambda+11=0 \\ & \Rightarrow(25 \lambda-11)(\lambda-1)=0 \\ & \Rightarrow \lambda=1, \frac{11}{25} \end{aligned}$

$\begin{aligned} & \overrightarrow{O P}=x \hat{i}+y \hat{j}+z \hat{k} \\ & \overrightarrow{O Q}=x \hat{i}+y \hat{j} \\ & |O P|=\gamma=\sqrt{x^2+y^2+z^2} \\ & \cos \theta=\frac{x}{\sqrt{x^2+y^2}} \Rightarrow \cos ^2 \theta=\frac{x^2}{\gamma^2-z^2}=\frac{x^2}{\gamma^2-\gamma^2 \cos ^2 \phi} \\ & \cos \phi=\frac{z}{\sqrt{x^2+y^2+z^2}}=\frac{z}{\gamma} \\ & \text { Distance of } P \text { from } x \text {-axis }=\sqrt{y^2+z^2} \\ & d=\sqrt{\gamma^2-x^2} \\ & \Rightarrow x^2=\gamma^2 \sin ^2 \phi \cos ^2 \theta \\ & \Rightarrow d=\sqrt{\gamma^2-\gamma^2 \sin ^2 \phi \cos ^2 \theta} \\ & =\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta} \end{aligned}$

$\begin{aligned} & L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k} \\ & L_1=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(\hat{i}-3 \hat{j}+4 \hat{k}) \\ & L_2: \vec{r}=2 \hat{i}+3 \hat{j}+5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k}) \\ & \vec{a}_1=2 \hat{i}+\hat{j}+3 \hat{k} \\ & \vec{a}_2=2 \hat{i}+3 \hat{j}+5 \hat{k} \\ & \vec{a}_2-\vec{a}_1=2 \hat{j}+2 \hat{k} \\ & \vec{b}_1=\hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}_2=2 \hat{i}+3 \hat{j}+\hat{k} \\ & \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 4 \\ 2 & 3 & 1 \end{array}\right| \end{aligned}$

$\begin{aligned} & \hat{i}(-3-12)-\hat{j}(1-8)+\hat{k}(3+6) \\ & =-15 \hat{i}+7 \hat{j}+9 \hat{k} \\ & \left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{225+49+81} \\ & \left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|=\frac{14+18}{\sqrt{355}}=\frac{32}{\sqrt{355}} \\ & m+n=387 \end{aligned}$

$\text { Line } L \Rightarrow \frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}=\mu$

Point $M(\mu, \mu+3,-\mu+1)$

Direction vector of $Q M=(\mu-3, \mu+6,-\mu)$

Direction vector of line $L=(1,1,-1)$

Both direction vector are perpendicular so

$\begin{aligned} & 1 \times(\mu-3)+1 \times(\mu+6)-1 \times(-\mu)=0 \\ & \mu=-1 \end{aligned}$

Point $M(-1,2,2)$ midpoint of $P Q$

So point $P(\alpha, \beta, \gamma)=(-5,7,3)$

Area of $\triangle P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{Q R}|$

$\begin{aligned} & \overrightarrow{P Q}=(8,-10,-2) \\ & \overrightarrow{Q R}=(-1,8,-2) \\ & \overrightarrow{P Q} \times \overrightarrow{Q R}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 8 & -10 & -2 \\ -1 & 8 & -2 \end{array}\right|=36 \hat{i}+18 \hat{j}+54 \hat{k} \end{aligned}$

Area of $\triangle P Q R=\lambda$

$\lambda=\frac{1}{2} \sqrt{(36)^2+(18)^2+(54)^2}$

$\begin{aligned} & \lambda^2=\frac{1}{4}(4536) \\ & \lambda^2=1134 \\ & \lambda^2=14 K \text { (given) } \\ & 14 K=1134 \\ & K=\frac{1134}{14}=81 \\ & K=81 \end{aligned}$

$A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1), D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$ are vertices of a quadrilateral

$\begin{aligned} & \overrightarrow{A C}=(2 \hat{i}+2 \hat{j}+\hat{k})-(3 \hat{i}+\hat{j}-\hat{k}) \\ & =-\hat{i}+\hat{j}+2 \hat{k} \\ & \overrightarrow{B D}=\left(\frac{10}{3} \hat{i}+\frac{2}{3} \hat{j}-\frac{1}{3} \hat{k}\right)-\left(\frac{5}{3} \hat{i}+\frac{7}{3} \hat{j}+\frac{1}{3} \hat{k}\right) \\ & \overrightarrow{B D}=\frac{5}{3} \hat{i}-\frac{5}{3} \hat{j}-\frac{2}{3} \hat{k} \\ & \text { Area }=\frac{1}{2}|\overrightarrow{A C} \times \overrightarrow{B D}| \\ & =\frac{1}{2}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ \frac{5}{3} & \frac{-5}{3} & \frac{-2}{3} \end{array}\right| \\ & =\frac{1}{2} \sqrt{\left(\frac{8}{3}\right)^2+\left(\frac{8}{3}\right)^2}\left[\because \overrightarrow{A C} \times \overrightarrow{B D}=\frac{8}{3} \hat{i}+\frac{8}{3} \hat{j}\right] \\ & =\frac{1}{2} \times \frac{8}{3} \times \sqrt{2}=\frac{4 \sqrt{2}}{3} \\ \end{aligned}$

Given two lines are represented as:

$ \frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5} $

and

$ \frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3} $

The formula for the shortest distance between two lines is:

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

From the given lines:

$ \vec{a}_1 = (3, -15, 9) $

$ \vec{a}_2 = (-1, 1, 9) $

$ \vec{b}_1 = (2, -7, 5) $

$ \vec{b}_2 = (2, 1, -3) $

Calculate the difference:

$ \vec{a}_2 - \vec{a}_1 = (-1-3, 1+15, 9-9) = (-4, 16, 0) $

Next, compute the cross product:

$ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = (16\hat{i} + 16\hat{j} + 16\hat{k}) $

The magnitude of the cross product is:

$ |\vec{b}_1 \times \vec{b}_2| = |(16\hat{i} + 16\hat{j} + 16\hat{k})| = 16\sqrt{3} $

Calculate the dot product:

$ (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-4\hat{i} + 16\hat{j} + 0\hat{k}) \cdot (16\hat{i} + 16\hat{j} + 16\hat{k}) = -64 + 256 + 0 = 192 $

Finally, find the shortest distance:

$ d = \frac{|192|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = 4\sqrt{3} $

Thus, the shortest distance is:

$ \boxed{4 \sqrt{3}} $

$(x, y, z) \equiv(2 \lambda+1,3 \lambda+1,5 \lambda+2)$

DR of $P Q$ :

$(2 \lambda-7,3 \lambda-6,5 \lambda-5)$

DR of line : $(2,3,5)$

$\begin{array}{ll} \therefore & 2(2 \lambda-7)+3(3 \lambda-6)+5(5 \lambda-5)=0 \\ \Rightarrow & \lambda=\frac{3}{2} \\ \therefore \quad & Q\left(4, \frac{7}{2}, \frac{19}{2}\right) \\ \therefore \quad & (\alpha, \beta, y) \equiv(0,2,12) \quad\left(Q \text { is mid point of } P \& P^{\prime}\right) \\ & \alpha+\beta+y \equiv 14 \end{array}$

$\begin{aligned} & L_1: \frac{x-2}{(-3)}=\frac{y-\frac{2}{3}}{\left(\frac{4 \lambda+1}{3}\right)}=\frac{}{(-1)} \\ & L_2: \frac{x+3}{3 \mu}=\frac{y-\frac{1}{2}}{-3}=\frac{z-5}{-7} \\ & \because L_1 \perp L_2 \\ & \Rightarrow(-3)(3 \mu)+\left(\frac{4 \lambda+1}{3}\right)(-3)+(-1)(-7)=0 \\ & -9 \mu-4 \lambda-1+7=0 \\ & \Rightarrow 4 \lambda+9 \mu=6 \\ & \end{aligned}$

$\begin{aligned} & P_1:(3 k-6,2 k, k-1) \\ & P_2(4 \alpha+7,3 \alpha+9,2 \alpha+4) \\ & P_1 \equiv P_2 \\ & 3 k-6=4 \alpha+7 \Rightarrow 3 k-4 \alpha=13 \\ & 2 k=3 \alpha+9 \Rightarrow 2 k-3 \alpha=9 \\ & \therefore k=3, \alpha=-1 \\ & \therefore P_1:(3,6,2) \end{aligned}$

Distance of $(3,6,2)$ and $(7,8,9)$

$\begin{aligned} & =\sqrt{16+4+49}=\sqrt{69}=d \\ & d^2+6=69+6=75 \end{aligned}$

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

Point of intersection of $L_1$ and $L_2$ is $(-1,1,-1)$

Distance of point $P$ from $L_3: 4 x=2 y=z$

$L_3: \frac{x}{\frac{1}{4}}=\frac{y}{\frac{1}{2}}=\frac{z}{1}$

Any point on $L_3$ be

$\begin{aligned} & \left(\frac{\lambda}{4}, \frac{\lambda}{2}, \lambda\right) \\ & P R:\left\langle\frac{\lambda}{4}+1, \frac{\lambda}{2}-1, \lambda+1\right\rangle \\ & \because P R \perp\left\langle\frac{1}{4}, \frac{1}{2}, 1\right\rangle \end{aligned}$

$\begin{aligned} & \Rightarrow\left(\frac{\lambda}{4}+1\right) \frac{1}{4}+\frac{1}{2}\left(\frac{\lambda}{2}-1\right)+\lambda+1=0 \\ & \quad \frac{\lambda}{16}+\frac{1}{4}+\frac{\lambda}{4}-\frac{1}{2}+\lambda+1=0 \\ & \Rightarrow \quad \lambda=\frac{-4}{7} \\ & \therefore \quad R\left(\frac{-1}{7}, \frac{-2}{7}, \frac{-4}{7}\right) \\ & \text { Now } R P: \sqrt{\left(\frac{-1}{7}+1\right)^2+\left(\frac{-2}{7}-1\right)^2+\left(\frac{-4}{7}+1\right)^2} \\ & \quad=\sqrt{\frac{36}{49}+\frac{81}{49}+\frac{9}{49}}=\frac{\sqrt{126}}{7}=\frac{3 \sqrt{14}}{7} \end{aligned}$

Line through $P Q$

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

Any point on $P Q$. be $R(4 \lambda+1,-2 \lambda-2,4 \lambda+3)$

$(P R)^2=81$

$(4 \lambda+1-1)^2+(-2 \lambda-2+2)^2+(4 \lambda+3-3)^2=81$

$16 \lambda^2+4 \lambda^2+16 \lambda^2=81$

$36 \lambda^2=81$

$\lambda= \pm \frac{9}{6}= \pm \frac{3}{2}$

$\therefore R$ can be $(7,-5,9)$ or $(-5,1,-3)$

Distance from origin for both points be $\sqrt{49+25+81}$ and $\sqrt{25+1+9}=\sqrt{35}$

$\therefore$ Distance of $(7,-5,9)$ is farthest from origin

$\therefore(\alpha, \beta, \gamma)=(7,-5,9)$

Now $7^2+(-5)^2+9^2=155$

Given the two lines:

$ L_1: \frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1} $

$ L_2: \frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1} $

We observe that these lines are not parallel as their directional vectors are not proportional. The directional vector for $L_1$ is $(-2,1,1)$ and for $L_2$ is $(1,-2,1)$. The shortest distance between two skew (non-intersecting and non-parallel) lines in the three-dimensional space is along the line that is perpendicular to both lines. This implies we can find a vector that is perpendicular to both directional vectors by taking their cross product.

The directional vector for $L_1$ is $d_1 = \langle -2, 1, 1 \rangle$, and for $L_2$ is $d_2 = \langle 1, -2, 1 \rangle$. The cross product of $d_1$ and $d_2$, which will be perpendicular to both lines, is given by:

$ d = d_1 \times d_2 = \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1 \\ \end{matrix} \right| $

Expanding the determinate gives:

$ d = \mathbf{i}((1)(1) - (1)(-2)) - \mathbf{j}((-2)(1) - (1)(1)) + \mathbf{k}((-2)(-2) - (1)(1)) $

$ d = \mathbf{i}(1 + 2) - \mathbf{j}(-2 - 1) + \mathbf{k}(4 - 1) $

$ d = 3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k} $

$ d = \langle 3, 3, 3 \rangle $

The shortest distance $D$ between the two lines can then be given by the formula:

$ D = \frac{\left| (\mathbf{a_2} - \mathbf{a_1}) \cdot d \right|}{\|d\|} $

Where $\mathbf{a_1}$ and $\mathbf{a_2}$ are position vectors to any points on line $L_1$ and line $L_2$, respectively, and '$\cdot$' denotes the dot product.

From the lines' equations, we can choose a point on each line (when the parameter is zero). Thus, for $L_1$, let's choose the point $A(\lambda, 2, 1)$, and for $L_2$, let's choose the point $B(\sqrt{3}, 1, 2)$. These points correspond to the vectors $\mathbf{a_1} = \langle \lambda, 2, 1 \rangle$ and $\mathbf{a_2} = \langle \sqrt{3}, 1, 2 \rangle$, respectively.

The vector $\mathbf{a_2} - \mathbf{a_1}$ is:

$ \mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3}, 1, 2 \rangle - \langle \lambda, 2, 1 \rangle $

$ \mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3} - \lambda, 1 - 2, 2 - 1 \rangle $

$ \mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3} - \lambda, -1, 1 \rangle $

We can now substitute this, along with $d$, into the distance formula:

$ D = \frac{\left| (\langle \sqrt{3} - \lambda, -1, 1 \rangle) \cdot \langle 3, 3, 3 \rangle \right|}{\|\langle 3, 3, 3 \rangle\|} $

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

$ D = \frac{\left| 3\sqrt{3} - 3\lambda - 3 + 3 \right|}{\sqrt{27}} $

$ D = \frac{\left| 3\sqrt{3} - 3\lambda \right|}{3\sqrt{3}} $

$ D = \frac{\left| \sqrt{3} - \lambda \right|}{\sqrt{3}} $

Given that the shortest distance $D$ between the lines is 1, we can equate the above result to 1, and solve for $\lambda$:

$ \frac{\left| \sqrt{3} - \lambda \right|}{\sqrt{3}} = 1 $

$ \left| \sqrt{3} - \lambda \right| = \sqrt{3} $

This absolute value equation gives us two possible cases:

Case 1: $\sqrt{3} - \lambda = \sqrt{3}$, which gives $\lambda = 0$.

Case 2: $\sqrt{3} - \lambda = -\sqrt{3}$, which gives $\lambda = 2\sqrt{3}$.

Therefore, the sum of all possible values of $\lambda$ is:

$ \lambda_{sum} = \lambda_1 + \lambda_2 = 0 + 2\sqrt{3} = 2\sqrt{3} $

Hence, option B ($2 \sqrt{3}$) is the correct answer.

$\begin{aligned} & \because \overrightarrow{\mathrm{PR}} \perp(2,3,4) \\ & \therefore \overrightarrow{\mathrm{PR}} \cdot(2,3,4)=0 \\ & (\alpha-2, \beta-3, \gamma-5) \cdot(2,3,4)=0 \\ & \Rightarrow 2 \alpha+3 \beta+4 \gamma=4+9+20=33 \end{aligned}$

$\begin{aligned} & \mathrm{L}_2=\frac{\mathrm{x}+4}{3}=\frac{\mathrm{y}-4}{2}=\frac{\mathrm{z}-3}{0} \\ & \therefore \mathrm{S} . \mathrm{D}=\frac{\left|\begin{array}{ccc} \mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{array}\right|}{\left|\overrightarrow{\mathrm{n}_1} \times \overrightarrow{\mathrm{n}_2}\right|} \\ & =\frac{\left|\begin{array}{ccc} 5 & -5 & -7 \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{array}\right|}{\left|\overrightarrow{\mathrm{n}_1} \times \overrightarrow{\mathrm{n}_2}\right|} \\ & =\frac{141}{|-4 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}|} \\ & =\frac{141}{\sqrt{16+36+169}} \\ & =\frac{141}{\sqrt{221}} \end{aligned}$

$\mathrm{L}_1 \perp \mathrm{L}_2 \quad \mathrm{~L}_3 \perp \mathrm{L}_1, \mathrm{~L}_2$

$\begin{aligned} & 3-1+2 \mathrm{P}=0 \\ & \mathrm{P}=-1 \\ & \left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 2 \\ 3 & 1 & -1 \end{array}\right|=-\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & \therefore(-\delta, 7 \delta, 4 \delta) \text { will lie on } \mathrm{L}_3 \end{aligned}$

For $\delta=1$ the point will be $(-1,7,4)$

Let foot $P(5 k-3,2 k+1,3 k-4)$

DR's $\rightarrow$ AP : $5 \mathrm{k}-4,2 \mathrm{k}-1,3 \mathrm{k}-7$

DR's $\rightarrow$ Line: $5,2,3$

Condition of perpendicular lines $(25 k-20)+(4 k-2)+(9 k-21)=0$

Then $\mathrm{k}=\frac{43}{38}$

Then $19(\alpha+\beta+\gamma)=\mathbf{1 0 1}$

$\begin{aligned} & \text { Area }=|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}| \\ & =\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 5 & -1 & 7 \\ 1 & 2 & 3 \end{array}\right| \\ & =\frac{1}{2}|-17 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}|=\frac{1}{2} \sqrt{474} \end{aligned}$

Direction ratio of $\mathrm{PR}=(4,1,-1)$

Direction ratio of $\mathrm{PQ}=(1,4,-1)$

Now, $\cos \theta=\left|\frac{4+4+1}{\sqrt{18} \cdot \sqrt{18}}\right|$

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

$\text { length of } O C=\frac{\sqrt{136}}{3}=\frac{2 \sqrt{34}}{3}$

Centroid $G$ divides MR in $1: 2$

$\mathrm{G}(1,2,2)$

Point of intersection $A$ of given lines is $(2,-6,0)$

$\mathrm{AG}=\sqrt{69}$

$\mathrm{L}_1=\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}=\lambda$

$\begin{aligned} & \mathrm{M}(\lambda, 1+2 \lambda, 2+3 \lambda) \\ & \overrightarrow{\mathrm{PM}}=(\lambda-1) \hat{\mathrm{i}}+(1+2 \lambda) \hat{\mathrm{j}}+(3 \lambda-5) \hat{\mathrm{k}} \end{aligned}$

$\overrightarrow{\mathrm{PM}}$ is perpendicular to line $\mathrm{L}_1$

$\begin{aligned} & \overrightarrow{\mathrm{PM}} \overrightarrow{\mathrm{b}}=0 \quad(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ & \Rightarrow \lambda-1+4 \lambda+2+9 \lambda-15=0 \\ & 14 \lambda=14 \Rightarrow \lambda=1 \\ & \therefore \mathrm{M}=(1,3,5) \\ & \overrightarrow{\mathrm{Q}}=2 \overrightarrow{\mathrm{M}}-\overrightarrow{\mathrm{P}}[\mathrm{M} \text { is midpoint of } \overrightarrow{\mathrm{P}} \& \overrightarrow{\mathrm{Q}}] \end{aligned}$

$\begin{aligned} & \overrightarrow{\mathrm{Q}}=2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}-\hat{\mathrm{i}}-7 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{Q}}=\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \therefore(\alpha, \beta, \gamma)=(1,6,3) \end{aligned}$

Required line having direction cosine $(l, \mathrm{~m}, \mathrm{n})$

$\begin{aligned} & l^2+m^2+n^2=1 \\ & \Rightarrow l^2+\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{\sqrt{2}}\right)^2=1 \end{aligned}$

$l^2=\frac{1}{4}$

$\therefore l=\frac{1}{2}$ [Line make acute angle with $\mathrm{x}$-axis]

Equation of line passing through $(1,6,3)$ will be

$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\mu\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}\right)$

Option (3) satisfying for $\mu=4$

$\mathrm{B}=(2 \lambda+7,-3 \lambda-2,6 \lambda+11)$

Point B lies on $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$

$\frac{2 \lambda+7-6}{1}=\frac{-3 \lambda-2-4}{0}=\frac{6 \lambda+11-8}{3}$

$-3 \lambda-6=0$

$\lambda=-2$

$\mathrm{B} \Rightarrow(3,4,-1)$

$\mathrm{AB}=\sqrt{(7-3)^2+(4+2)^2+(11+1)^2}$

$=\sqrt{16+36+144}$

$=\sqrt{196}=14$

$\begin{aligned} & \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3} \\ & \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5} \end{aligned}$

the shortest distance between the lines

$=\left|\frac{(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot\left(\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right)}{\left|\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right|}\right|$

$=\left|\frac{\left|\begin{array}{ccc}\lambda-4 & 0 & 2 \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|}\right|$

$=\left|\frac{(\lambda-4)(-10+12)-0+2(4-4)}{|2 \hat{i}-1 \hat{j}+0 \hat{k}|}\right|$

$\begin{aligned} & \frac{6}{\sqrt{5}}=\left|\frac{2(\lambda-4)}{\sqrt{5}}\right| \\ & 3=|\lambda-4| \\ & \lambda-4= \pm 3 \\ & \lambda=7,1 \end{aligned}$

Sum of all possible values of $\lambda$ is $=8$