3D Geometry

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 $=8$

$\begin{aligned} & L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}=a \\ & L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}=b \end{aligned}$

$\begin{aligned} & x=a-11=3 b-16 \Rightarrow a-3 b=-5 \quad \text{.... (1)}\\ & y=2 a-21=2 b-11 \Rightarrow 2 a-2 b=10 \quad \text{.... (2)}\\ & z=3 a-29=b r-4 \Rightarrow 3 a-b y=25 \quad \text{.... (3)} \end{aligned}$

$\begin{aligned} & \text { from (1) & (2) } \\ & a=10, b=5 \end{aligned}$

Now from (3)

$\begin{aligned} & 3(10)-5 \gamma=25 \quad \therefore \gamma=1 \\ & \mathrm{R}_1 \equiv(-1,-1,1) \\ & \mathrm{OR}_1=-\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} \end{aligned}$

$\begin{aligned} & \vec{n}=\left|\begin{array}{lll} i & j & k \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right|=-4 \hat{i}-(-8) \hat{j}-4 \hat{k} \\ & \vec{n}=-4 \hat{i}+8 \hat{j}+4 \hat{k}=-4(\hat{i}-2 \hat{j}+\hat{k}) \\ & \hat{n}= \pm \frac{4(\hat{i}-2 \hat{j}+\hat{k})}{4 \sqrt{6}}= \pm \frac{(\hat{i}-2 \hat{j}+\hat{k})}{\sqrt{6}} \\ & \overrightarrow{O R} \cdot \hat{n}= \pm(-\hat{i}-\hat{j}+\hat{k})\left(\frac{\hat{i}-2 \hat{j}+\hat{k}}{\sqrt{6}}\right)= \pm \frac{2}{\sqrt{6}}= \pm \sqrt{\frac{4}{6}}= \pm \sqrt{\frac{2}{3}} \end{aligned}$

$\begin{aligned} & \text { line }: \frac{x-1}{1}=\frac{y-3}{2}=\frac{z-2}{1} \\ & (x, y, z)=(\lambda+1,2 \lambda+3, \lambda+2) \\ & \text { Put in } L_1: x-y+3 z=6 \\ & (\lambda+1)-(2 \lambda+3)+3(\lambda+2)=6 \\ & 2 \lambda=2 \Rightarrow \lambda=1 \\ & Q=(2,5,3) \end{aligned}$

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

$(\mathrm{x}, \mathrm{y}, \mathrm{z})=(\mathrm{t}+2,5-\mathrm{t}, 3 \mathrm{t}+3)$

Put in $L_2: 2 x-y+z=-4$

$\begin{aligned} & 2(t+4)-(5-t)+(3 t+3)=-4 \\ & 6 t=-6 \Rightarrow t=-1 \\ & R=(1,6,0) \end{aligned}$

$\begin{aligned} & \text { Perimeter }=\sqrt{6}+\sqrt{13}+\sqrt{11} \\ & \text { Centroid }=\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right) \end{aligned}$

$\begin{aligned} & \mathrm{S}=\left\{\mathrm{X}:(\mathrm{XP})^2-(\mathrm{XQ})^2=50\right\} \\ & \mathrm{T}=\left\{\mathrm{Y}:(\mathrm{YQ})^2-(\mathrm{YP})^2=50\right\} \end{aligned}$

for finding $S \equiv X(x, y, z)$ and for $T \equiv Y(x, y, z)$

$\begin{aligned} & \left((x-1)^2+(y-1)^2+(z-1)^2\right)-\left((x-4)^2+(y-2)^2+(z-7)^2\right)=50 \\ & \Rightarrow \quad S=\{(x, y, z): 6 x+8 z=105\} \\ & \quad \quad T=\{(x, y, z): 6 x+8 z=5\} \end{aligned}$

Since S and T both are plane;

(A) There exist a triangle in plane S whose area $=1$ (always)

(B) L & M lies on plane T, hence line segment joining L & M will lie on plane T.

(C) Distance between $\mathrm{S}$ & $\mathrm{T}$

$\mathrm{d}=\left|\frac{105-5}{10}\right|=10$

Hence for rectangle of perimeter 48 can exist.

There will be infinite such rectangle possible.

(D) For square

Hence Answers A, B, C, D are correct.

The equation of the plane passing through points $A(-1, -2, -3)$, $B(9, 3, 4)$, and $C(9, -2, 1)$ can be written using the determinant :

$ \left|\begin{array}{ccc} x+1 & y+2 & z+3 \\ 10 & 5 & 7 \\ 10 & 0 & 4 \end{array}\right|=0 $

Expanding the determinant, we get :

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

Next, we find the foot of the perpendicular from point $P(3, -2, -9)$ to the plane. Using the coordinates of $P$ and the equation of the plane, we can find the ratio of the perpendicular distance to the sum of the squares of the coefficients of $x$, $y$, and $z$ :

$ \frac{2(3) + 3(-2) - 5(-9) - 7}{2^2 + 3^2 + (-5)^2} = \frac{-38}{38} $

We can now find the coordinates of the foot of the perpendicular, $Q(\alpha, \beta, \gamma)$ :

$ \frac{\alpha - 3}{2} = \frac{\beta + 2}{3} = \frac{\gamma + 9}{-5} = -\frac{38}{38} $

Solving for $\alpha$, $\beta$, and $\gamma$ :

$ \alpha = 3 - 2\left(-\frac{38}{38}\right) = 1 $

$ \beta = -2 + 3\left(-\frac{38}{38}\right) = -5 $

$ \gamma = -9 - 5\left(-\frac{38}{38}\right) = -4 $

So, the coordinates of the foot of the perpendicular are $Q(1, -5, -4)$. Now, we can find the distance of point $Q$ from the origin :

$ OQ = \sqrt{\alpha^2 + \beta^2 + \gamma^2} = \sqrt{1^2 + (-5)^2 + (-4)^2} = \sqrt{42} $

Given two lines:

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

and

$\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}$

These lines are coplanar if the determinant of the matrix

$ \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} $ = 0

Now let's apply this condition to the given problem. The given line is :

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

So, the coordinates of any point on this line are $(-3, 1, 5)$ and the direction ratios are $(-3, 1, 5)$.

Now, let's calculate the determinants for each option and check which one equals zero.

For Option A :

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

The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(-1, 2, 4)$.

The determinant is :

$ \begin{vmatrix} -1 - (-3) & 2 - 1 & 5 - 5 \\ -3 & 1 & 5 \\ -1 & 2 & 4 \end{vmatrix} = \begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 4 \end{vmatrix} $

Applying the formula :

= $2(1 \times 4 - 5\times2) - 1((-3\times4) - (5\times-1)) + 0((-3\times2) - (1\times-1))$

= $2(4 - 10) - 1(-12 - (-5)) + 0(-6 - (-1))$

= $2(-6) - 1(-7) + 0(-5)$

= $-12 + 7 + 0$

= $-5$

The determinant for Option A is not equal to zero, so this line is not coplanar with the given line.

For Option B, we have :

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

The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(-1, 2, 5)$.

$ \begin{vmatrix} -1 - (-3) & 2 - 1 & 5 - 5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix} = \begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix} $

Applying the formula, we get :

= $2(1*5 - 5*2) - 1((-3*5) - (5*-1)) + 0((-3*2) - (1*-1))$

= $2(5 - 10) - 1(-15 - (-5)) + 0(-6 - (-1))$

= $2(-5) - 1(-10) + 0(-5)$

= $-10 + 10 + 0$

= $0$

For Option C :

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

The coordinates of any point on this line are $(1, 2, 5)$ and the direction ratios are $(-1, 2, 5)$.

The determinant is:

$ \begin{vmatrix} 1 - (-3) & 2 - 1 & 5 - 5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix} = \begin{vmatrix} 4 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix} $

Applying the formula :

= $4(1*5 - 5*2) - 1((-3*5) - (5*-1)) + 0((-3*2) - (1*-1))$

= $4(5 - 10) - 1(-15 - (-5)) + 0(-6 - (-1))$

= $4(-5) - 1(-10) + 0(-5)$

= $-20 + 10 + 0$

= $-10$

The determinant for Option C is not equal to zero, so this line is not coplanar with the given line.

For Option D :

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

The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(1, 2, 5)$.

The determinant is :

$ \begin{vmatrix} -1 - (-3) & 2 - 1 & 5 - 5 \\ -3 & 1 & 5 \\ 1 & 2 & 5 \end{vmatrix} = \begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ 1 & 2 & 5 \end{vmatrix} $

Applying the formula :

= $2(1*5 - 5*2) - 1((-3*5) - (5*1)) + 0((-3*2) - (1*1))$

= $2(5 - 10) - 1(-15 - 5) + 0(-6 - 1)$

= $2(-5) - 1(-20) + 0(-7)$

= $-10 + 20 + 0$

= $10$

So the determinant for Option D is not equal to zero, which means the line in Option D is not coplanar with the given line.

The first step is to find the normal vector to the desired plane. Since the plane is parallel to the line passing through the points (5, 1, -7) and (1, -1, -1), the direction vector of that line is also parallel to the plane. The direction vector is the difference between the coordinates of the two points, which is (5-1, 1-(-1), -7-(-1)) = (4, 2, -6).

Next, let's find another vector that is parallel to the plane. This vector can be obtained by taking the difference between the coordinates of the points (0, -1, 2) and (-1, 2, 1), which the plane passes through. This gives us a vector of (0-(-1), -1-2, 2-1) = (1, -3, 1).

The normal to the plane is perpendicular to both these vectors. It can be found by taking the cross product of the two vectors.

The cross product of vectors (4, 2, -6) and (1, -3, 1) is :

The equation of the plane can now be written in the form :

-16x - 10y - 14z = d

We can find the constant 'd' by substituting one of the points through which the plane passes, say (0, -1, 2) :

d = -16.0 -10.(-1) -14.2 = 10 - 28 = -18

So, the equation of the plane is -16x - 10y - 14z = -18.

Now, we substitute the given options into the equation to check which one satisfies it :

Option A: (-16.0 -10.5 -14.(-2) = -18) => -50 + 28 = -22 ≠ -18, so A is not correct.

Option B: (-16.2 -10.0 -14.1 = -18) => -32 - 14 = -46 ≠ -18, so B is not correct.

Option C: (-16.1 -10.(-2) -14.1 = -18) => -16 + 20 -14 = -10 ≠ -18, so C is not correct.

Option D: (-16.(-2) -10.5 -14.0 = -18) => 32 - 50 = -18, so D is correct.

So, the plane also passes through the point given in Option D, which is (-2, 5, 0).