Line Questions in English

(A) The direction ratios of line $L$ passing through $P(1, 2, 1)$ and $Q(2, 1, -1)$ are $(2-1, 1-2, -1-1) = (1, -1, -2)$.
The equation of line $L$ is $\frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{-2} = k$.
Any point $C$ on line $L$ is $(k+1, -k+2, -2k+1)$.
Since $AC$ is perpendicular to line $L$,the direction ratios of $AC$ are $(k+1-2, -k+2-2, -2k+1-2) = (k-1, -k, -2k-1)$.
Since $AC \perp L$,the dot product of their direction ratios is zero:
$1(k-1) + (-1)(-k) + (-2)(-2k-1) = 0$
$k - 1 + k + 4k + 2 = 0$
$6k + 1 = 0 \Rightarrow k = -\frac{1}{6}$.
Substituting $k$ in the coordinates of $C$,we get $C = (1 - \frac{1}{6}, 2 + \frac{1}{6}, 1 + \frac{2}{6}) = (\frac{5}{6}, \frac{13}{6}, \frac{8}{6})$.
Since $C$ is the midpoint of $AB$,where $B = (\alpha, \beta, \gamma)$ and $A = (2, 2, 2)$:
$\frac{\alpha+2}{2} = \frac{5}{6} \Rightarrow \alpha+2 = \frac{5}{3} \Rightarrow \alpha = -\frac{1}{3}$.
$\frac{\beta+2}{2} = \frac{13}{6} \Rightarrow \beta+2 = \frac{13}{3} \Rightarrow \beta = \frac{7}{3}$.
$\frac{\gamma+2}{2} = \frac{8}{6} \Rightarrow \gamma+2 = \frac{8}{3} \Rightarrow \gamma = \frac{2}{3}$.
Now,$\alpha + \beta + 6\gamma = -\frac{1}{3} + \frac{7}{3} + 6(\frac{2}{3}) = \frac{6}{3} + 4 = 2 + 4 = 6$.

(C) Let the first line be $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=\lambda$. The general point on this line is $(3\lambda-6, 2\lambda, \lambda-1)$.
Let the second line be $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=\mu$. The general point on this line is $(4\mu+7, 3\mu+9, 2\mu+4)$.
Equating the coordinates for the point of intersection:
$3\lambda-6 = 4\mu+7 \Rightarrow 3\lambda-4\mu = 13$ (Equation $1$)
$2\lambda = 3\mu+9 \Rightarrow 2\lambda-3\mu = 9$ (Equation $2$)
Solving these equations:
Multiply Equation $1$ by $2$ and Equation $2$ by $3$:
$6\lambda-8\mu = 26$
$6\lambda-9\mu = 27$
Subtracting the two equations gives $\mu = -1$.
Substituting $\mu = -1$ into Equation $2$: $2\lambda - 3(-1) = 9 \Rightarrow 2\lambda = 6 \Rightarrow \lambda = 3$.
The point of intersection is $(3(3)-6, 2(3), 3-1) = (3, 6, 2)$.
The distance $d$ from $(3, 6, 2)$ to $(7, 8, 9)$ is given by:
$d^2 = (7-3)^2 + (8-6)^2 + (9-2)^2 = 4^2 + 2^2 + 7^2 = 16 + 4 + 49 = 69$.
Therefore,$d^2+6 = 69+6 = 75$.

(D) The given lines are $\frac{2-x}{3}=\frac{3y-2}{4\lambda+1}=4-z$ and $\frac{x+3}{3\mu}=\frac{1-2y}{6}=\frac{5-z}{7}$.
First,rewrite the lines in standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$.
For the first line: $\frac{x-2}{-3}=\frac{y-2/3}{(4\lambda+1)/3}=\frac{z-4}{-1}$. The direction vector is $\vec{v_1} = (-3, \frac{4\lambda+1}{3}, -1)$.
For the second line: $\frac{x+3}{3\mu}=\frac{y-1/2}{-3}=\frac{z-5}{-7}$. The direction vector is $\vec{v_2} = (3\mu, -3, -7)$.
Since the lines are perpendicular,their dot product is zero: $\vec{v_1} \cdot \vec{v_2} = 0$.
$(-3)(3\mu) + (\frac{4\lambda+1}{3})(-3) + (-1)(-7) = 0$.
$-9\mu - (4\lambda+1) + 7 = 0$.
$-9\mu - 4\lambda - 1 + 7 = 0$.
$-4\lambda - 9\mu + 6 = 0$.
$4\lambda + 9\mu = 6$.

(C) Let the line be $L: \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5} = \lambda$. Any point $M$ on the line is given by $M(2\lambda+1, 3\lambda-1, 5\lambda+2)$.
Since $M$ is the foot of the perpendicular from $A(8, 5, 7)$ to the line,the vector $\overrightarrow{AM}$ is perpendicular to the direction vector of the line $\vec{v} = 2\hat{i} + 3\hat{j} + 5\hat{k}$.
$\overrightarrow{AM} = (2\lambda+1-8)\hat{i} + (3\lambda-1-5)\hat{j} + (5\lambda+2-7)\hat{k} = (2\lambda-7)\hat{i} + (3\lambda-6)\hat{j} + (5\lambda-5)\hat{k}$.
Since $\overrightarrow{AM} \cdot \vec{v} = 0$,we have:
$2(2\lambda-7) + 3(3\lambda-6) + 5(5\lambda-5) = 0$
$4\lambda - 14 + 9\lambda - 18 + 25\lambda - 25 = 0$
$38\lambda - 57 = 0 \implies \lambda = \frac{57}{38} = \frac{3}{2}$.
Substituting $\lambda = \frac{3}{2}$ into $M$,we get $M(2(\frac{3}{2})+1, 3(\frac{3}{2})-1, 5(\frac{3}{2})+2) = M(4, \frac{7}{2}, \frac{19}{2})$.
Let $A'(\alpha, \beta, \gamma)$ be the image of $A$. Since $M$ is the midpoint of $AA'$,we have:
$\frac{\alpha+8}{2} = 4 \implies \alpha = 0$
$\frac{\beta+5}{2} = \frac{7}{2} \implies \beta = 2$
$\frac{\gamma+7}{2} = \frac{19}{2} \implies \gamma = 12$
Thus,$\alpha + \beta + \gamma = 0 + 2 + 12 = 14$.

(D) Let the two lines be $L_1: \frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2} = \lambda$ and $L_2: \frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0} = \mu$.
Any point on $L_1$ is $P(-3\lambda-2, 4\lambda+2, 2\lambda+5)$ and any point on $L_2$ is $Q(-\mu-2, 2\mu-6, 1)$.
The direction ratios of the line of shortest distance $PQ$ are proportional to the cross product of the direction vectors of $L_1$ and $L_2$,which are $\vec{v_1} = -3\hat{i} + 4\hat{j} + 2\hat{k}$ and $\vec{v_2} = -1\hat{i} + 2\hat{j} + 0\hat{k}$.
$\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 4 & 2 \\ -1 & 2 & 0 \end{vmatrix} = \hat{i}(0-4) - \hat{j}(0 - (-2)) + \hat{k}(-6 - (-4)) = -4\hat{i} - 2\hat{j} - 2\hat{k}$.
This is parallel to the vector $2\hat{i} + \hat{j} + \hat{k}$.
The vector $\vec{PQ} = ((-3\lambda-2) - (-\mu-2))\hat{i} + ((4\lambda+2) - (2\mu-6))\hat{j} + ((2\lambda+5) - 1)\hat{k} = (\mu-3\lambda)\hat{i} + (4\lambda-2\mu+8)\hat{j} + (2\lambda+4)\hat{k}$.
Since $\vec{PQ}$ is parallel to $2\hat{i} + \hat{j} + \hat{k}$,we have $\frac{\mu-3\lambda}{2} = \frac{4\lambda-2\mu+8}{1} = \frac{2\lambda+4}{1}$.
From $\frac{4\lambda-2\mu+8}{1} = \frac{2\lambda+4}{1}$,we get $2\lambda - 2\mu + 4 = 0 \Rightarrow \mu = \lambda + 2$.
Substituting $\mu = \lambda + 2$ into $\frac{\mu-3\lambda}{2} = 2\lambda+4$,we get $\frac{\lambda+2-3\lambda}{2} = 2\lambda+4 \Rightarrow -\lambda+1 = 2\lambda+4 \Rightarrow 3\lambda = -3 \Rightarrow \lambda = -1$.
Then $\mu = -1+2 = 1$.
The line of shortest distance passes through $P(-3(-1)-2, 4(-1)+2, 2(-1)+5) = P(1, -2, 3)$ and $Q(-1-2, 2(1)-6, 1) = Q(-3, -4, 1)$.
The equation of the line $PQ$ is $\frac{x-1}{-3-1} = \frac{y-(-2)}{-4-(-2)} = \frac{z-3}{1-3} \Rightarrow \frac{x-1}{-4} = \frac{y+2}{-2} = \frac{z-3}{-2} \Rightarrow \frac{x-1}{2} = \frac{y+2}{1} = \frac{z-3}{1}$.
Since $(-1, \alpha, \beta)$ lies on this line,$\frac{-1-1}{2} = \frac{\alpha+2}{1} = \frac{\beta-3}{1} \Rightarrow -1 = \alpha+2 = \beta-3$.
Thus,$\alpha = -3$ and $\beta = 2$.
Therefore,$(\alpha-\beta)^2 = (-3-2)^2 = (-5)^2 = 25$.

(B) The given lines are $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$ and $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$.
The formula for the shortest distance $(S.D.)$ between two lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is $S.D. = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.
From the equations,we have:
$\vec{a}_1 = (3, -15, 9)$,$\vec{b}_1 = (2, -7, 5)$
$\vec{a}_2 = (-1, 1, 9)$,$\vec{b}_2 = (2, 1, -3)$
Calculate $\vec{a}_2 - \vec{a}_1 = (-1-3, 1-(-15), 9-9) = (-4, 16, 0)$.
Calculate the cross product $\vec{b}_1 \times \vec{b}_2$:
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = \hat{i}(21-5) - \hat{j}(-6-10) + \hat{k}(2+14) = 16\hat{i} + 16\hat{j} + 16\hat{k} = 16(\hat{i} + \hat{j} + \hat{k})$.
The magnitude $|\vec{b}_1 \times \vec{b}_2| = 16 \sqrt{1^2 + 1^2 + 1^2} = 16 \sqrt{3}$.
Now,calculate the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-4, 16, 0) \cdot (16, 16, 16) = -64 + 256 + 0 = 192$.
Therefore,$S.D. = \frac{|192|}{16 \sqrt{3}} = \frac{12}{\sqrt{3}} = 4 \sqrt{3}$.

(A) The line passes through $A(2, -5, 11)$ and $B(-6, 7, -5)$.
The direction ratios of the line are $(-6-2, 7-(-5), -5-11) = (-8, 12, -16)$.
Dividing by $-4$,the simplified direction ratios are $(2, -3, 4)$.
The equation of the line is $\frac{x-2}{2} = \frac{y+5}{-3} = \frac{z-11}{4} = \lambda$.
Any point $Q$ on the line is $(2\lambda+2, -3\lambda-5, 4\lambda+11)$.
The vector $\vec{RQ} = (2\lambda+2-1, -3\lambda-5-7, 4\lambda+11-6) = (2\lambda+1, -3\lambda-12, 4\lambda+5)$.
Since $RQ$ is perpendicular to the line,the dot product of $\vec{RQ}$ and the direction vector $(2, -3, 4)$ is $0$:
$2(2\lambda+1) - 3(-3\lambda-12) + 4(4\lambda+5) = 0$
$4\lambda + 2 + 9\lambda + 36 + 16\lambda + 20 = 0$
$29\lambda + 58 = 0 \Rightarrow \lambda = -2$.
Substituting $\lambda = -2$ into the coordinates of $Q$:
$Q = (2(-2)+2, -3(-2)-5, 4(-2)+11) = (-2, 1, 3)$.
The length $PQ$ where $P(10, -2, -1)$ and $Q(-2, 1, 3)$ is:
$PQ = \sqrt{(-2-10)^2 + (1-(-2))^2 + (3-(-1))^2}$
$PQ = \sqrt{(-12)^2 + (3)^2 + (4)^2} = \sqrt{144 + 9 + 16} = \sqrt{169} = 13$.

(D) Let the line be $L: \frac{x}{1} = \frac{y-3}{1} = \frac{z-1}{-1} = t$. Any point on the line is $S(t, t+3, 1-t)$.
Vector $\vec{QS} = (t-3, t+6, -t)$. The direction vector of the line is $\vec{v} = (1, 1, -1)$.
Since $\vec{QS} \perp \vec{v}$,we have $(t-3)(1) + (t+6)(1) + (-t)(-1) = 0$,which gives $t-3+t+6+t = 0 \implies 3t+3=0 \implies t=-1$.
Thus,the foot of the perpendicular is $S(-1, 2, 2)$.
Since $S$ is the midpoint of $PQ$,we have $\frac{\alpha+3}{2} = -1, \frac{\beta-3}{2} = 2, \frac{\gamma+1}{2} = 2$,so $P(-5, 7, 3)$.
Vector $\vec{RQ} = (3-2, -3-5, 1-(-1)) = (1, -8, 2)$ and $\vec{RP} = (-5-2, 7-5, 3-(-1)) = (-7, 2, 4)$.
The area of $\triangle PQR = \frac{1}{2} |\vec{RQ} \times \vec{RP}|$.
$\vec{RQ} \times \vec{RP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -8 & 2 \\ -7 & 2 & 4 \end{vmatrix} = \hat{i}(-32-4) - \hat{j}(4+14) + \hat{k}(2-56) = -36\hat{i} - 18\hat{j} - 54\hat{k}$.
Area $\lambda = \frac{1}{2} \sqrt{(-36)^2 + (-18)^2 + (-54)^2} = \frac{1}{2} \sqrt{1296 + 324 + 2916} = \frac{1}{2} \sqrt{4536} = \frac{1}{2} \sqrt{324 \times 14} = \frac{18}{2} \sqrt{14} = 9\sqrt{14}$.
Given $\lambda^2 = 14K$,we have $(9\sqrt{14})^2 = 81 \times 14 = 14K$,so $K = 81$.

(C) The lines are given by $\vec{r} = \vec{a}_1 + t\vec{p}$ and $\vec{r} = \vec{a}_2 + s\vec{q}$,where $\vec{a}_1 = \lambda \hat{i} + 2 \hat{j} + \hat{k}$,$\vec{p} = 3 \hat{i} - \hat{j} + \hat{k}$,$\vec{a}_2 = -2 \hat{i} - 5 \hat{j} + 4 \hat{k}$,and $\vec{q} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$.
The shortest distance $d$ is given by $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{p} \times \vec{q})|}{||\vec{p} \times \vec{q}||}$.
First,calculate $\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} = \hat{i}(-4-2) - \hat{j}(12+3) + \hat{k}(6-3) = -6 \hat{i} - 15 \hat{j} + 3 \hat{k}$.
The magnitude $||\vec{p} \times \vec{q}|| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}$.
Now,$\vec{a}_2 - \vec{a}_1 = (-2-\lambda) \hat{i} - 7 \hat{j} + 3 \hat{k}$.
$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{p} \times \vec{q}) = (-2-\lambda)(-6) + (-7)(-15) + (3)(3) = 12 + 6\lambda + 105 + 9 = 6\lambda + 126$.
Given $d = \frac{44}{\sqrt{30}}$,we have $\frac{|6\lambda + 126|}{3\sqrt{30}} = \frac{44}{\sqrt{30}}$.
$|6\lambda + 126| = 132$.
This implies $6\lambda + 126 = 132$ or $6\lambda + 126 = -132$.
Case $1$: $6\lambda = 6 \implies \lambda = 1$.
Case $2$: $6\lambda = -258 \implies \lambda = -43$.
The possible values of $|\lambda|$ are $|1| = 1$ and $|-43| = 43$.
Thus,the largest possible value of $|\lambda|$ is $43$.

(B) The lines are given by:
$L_1: \overrightarrow{r} = (2\hat{i} + \hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 4\hat{k})$
$L_2: \overrightarrow{r} = (2\hat{i} + 3\hat{j} + 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})$
Let $\overrightarrow{a_1} = 2\hat{i} + \hat{j} + 3\hat{k}$ and $\overrightarrow{a_2} = 2\hat{i} + 3\hat{j} + 5\hat{k}$.
Let $\overrightarrow{p} = \hat{i} - 3\hat{j} + 4\hat{k}$ and $\overrightarrow{q} = 2\hat{i} + 3\hat{j} + \hat{k}$.
Then $\overrightarrow{a_2} - \overrightarrow{a_1} = (2-2)\hat{i} + (3-1)\hat{j} + (5-3)\hat{k} = 0\hat{i} + 2\hat{j} + 2\hat{k}$.
The cross product $\overrightarrow{p} \times \overrightarrow{q}$ is:
$\overrightarrow{p} \times \overrightarrow{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 4 \\ 2 & 3 & 1 \end{vmatrix} = \hat{i}(-3-12) - \hat{j}(1-8) + \hat{k}(3+6) = -15\hat{i} + 7\hat{j} + 9\hat{k}$.
The magnitude $|\overrightarrow{p} \times \overrightarrow{q}| = \sqrt{(-15)^2 + 7^2 + 9^2} = \sqrt{225 + 49 + 81} = \sqrt{355}$.
The shortest distance is given by $d = \left| \frac{(\overrightarrow{a_2} - \overrightarrow{a_1}) \cdot (\overrightarrow{p} \times \overrightarrow{q})}{|\overrightarrow{p} \times \overrightarrow{q}|} \right|$.
$d = \left| \frac{(0\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (-15\hat{i} + 7\hat{j} + 9\hat{k})}{\sqrt{355}} \right| = \left| \frac{0 + 14 + 18}{\sqrt{355}} \right| = \frac{32}{\sqrt{355}}$.
Comparing with $\frac{m}{\sqrt{n}}$,we have $m = 32$ and $n = 355$.
Since $\operatorname{gcd}(32, 355) = 1$,the value of $m+n = 32 + 355 = 387$.

(C) The given lines are $L_1: \frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $L_2: \frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$.
Note that the direction vectors are $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b}_2 = 4\hat{i} + 6\hat{j} + 8\hat{k} = 2(2\hat{i} + 3\hat{j} + 4\hat{k})$.
Since $\vec{b}_2 = 2\vec{b}_1$,the lines are parallel.
The shortest distance between two parallel lines $\vec{r} = \vec{a}_1 + t\vec{b}$ and $\vec{r} = \vec{a}_2 + s\vec{b}$ is given by $d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$.
Here,$\vec{a}_1 = \lambda\hat{i} + 4\hat{j} + 3\hat{k}$,$\vec{a}_2 = 2\hat{i} + 4\hat{j} + 7\hat{k}$,and $\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
$\vec{a}_2 - \vec{a}_1 = (2-\lambda)\hat{i} + 0\hat{j} + 4\hat{k}$.
$(\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2-\lambda & 0 & 4 \\ 2 & 3 & 4 \end{vmatrix} = \hat{i}(0-12) - \hat{j}(8-12+6\lambda) + \hat{k}(6\lambda-6) = -12\hat{i} + (4-6\lambda)\hat{j} + (6\lambda-6)\hat{k}$.
Magnitude $|(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{(-12)^2 + (4-6\lambda)^2 + (6\lambda-6)^2} = \sqrt{144 + 16 - 48\lambda + 36\lambda^2 + 36\lambda^2 - 72\lambda + 36} = \sqrt{72\lambda^2 - 120\lambda + 196}$.
$|\vec{b}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4+9+16} = \sqrt{29}$.
Given $d = \frac{13}{\sqrt{29}}$,so $\frac{\sqrt{72\lambda^2 - 120\lambda + 196}}{\sqrt{29}} = \frac{13}{\sqrt{29}}$.
$72\lambda^2 - 120\lambda + 196 = 169 \implies 72\lambda^2 - 120\lambda + 27 = 0$.
Dividing by $3$,$24\lambda^2 - 40\lambda + 9 = 0$.
Using the quadratic formula $\lambda = \frac{40 \pm \sqrt{1600 - 864}}{48} = \frac{40 \pm \sqrt{736}}{48}$.
Wait,checking the calculation: $|12\hat{i} - 4\lambda\hat{j} + (3\lambda-6)\hat{k}| = 13 \implies 144 + 16\lambda^2 + 9\lambda^2 - 36\lambda + 36 = 169 \implies 25\lambda^2 - 36\lambda + 11 = 0$.
$(25\lambda - 11)(\lambda - 1) = 0$. Thus $\lambda = 1$ or $\lambda = 11/25$.

(D) Let the line be $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} = t$. Any point $A$ on the line is given by $A(t, 2t+1, 3t+2)$.
Since $A$ is the foot of the perpendicular from $Q(1, 6, 4)$ to the line,the vector $\overrightarrow{QA} = (t-1)\hat{i} + (2t-5)\hat{j} + (3t-2)\hat{k}$ must be perpendicular to the direction vector of the line $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.
Thus,$\overrightarrow{QA} \cdot \vec{b} = 0 \implies 1(t-1) + 2(2t-5) + 3(3t-2) = 0$.
$t - 1 + 4t - 10 + 9t - 6 = 0 \implies 14t - 17 = 0 \implies t = \frac{17}{14}$.
The foot of the perpendicular $A$ is $(\frac{17}{14}, 2(\frac{17}{14})+1, 3(\frac{17}{14})+2) = (\frac{17}{14}, \frac{48}{14}, \frac{79}{14})$.
Since $A$ is the midpoint of $PQ$,where $P(\alpha, \beta, \gamma)$ is the image of $Q(1, 6, 4)$,we have $\frac{\alpha+1}{2} = \frac{17}{14}$,$\frac{\beta+6}{2} = \frac{48}{14}$,and $\frac{\gamma+4}{2} = \frac{79}{14}$.
$\alpha = \frac{17}{7} - 1 = \frac{10}{7}$,$\beta = \frac{48}{7} - 6 = \frac{6}{7}$,$\gamma = \frac{79}{7} - 4 = \frac{51}{7}$.
Then $2\alpha + \beta + \gamma = 2(\frac{10}{7}) + \frac{6}{7} + \frac{51}{7} = \frac{20+6+51}{7} = \frac{77}{7} = 11$.

(B) Let the line $L$ intersect the line $L_1: \frac{x-2}{1} = \frac{y}{-1} = \frac{z-1}{1} = \lambda$ at point $M(2+\lambda, -\lambda, 1+\lambda)$ and the line $L_2: \frac{x+1}{1/2} = \frac{y-1}{1/2} = \frac{z+1}{1} = \mu$ at point $N(-1+\mu/2, 1+\mu/2, -1+\mu)$.
Since $L$ is parallel to the line with direction ratios $\langle 3, 1, 2 \rangle$,the vector $\vec{MN}$ must be parallel to $\langle 3, 1, 2 \rangle$.
$\vec{MN} = \langle (-1+\mu/2) - (2+\lambda), (1+\mu/2) - (-\lambda), (-1+\mu) - (1+\lambda) \rangle = \langle \mu/2 - \lambda - 3, \mu/2 + \lambda + 1, \mu - \lambda - 2 \rangle$.
Since $\vec{MN} \parallel \langle 3, 1, 2 \rangle$,we have $\frac{\mu/2 - \lambda - 3}{3} = \frac{\mu/2 + \lambda + 1}{1} = \frac{\mu - \lambda - 2}{2}$.
From $\frac{\mu/2 + \lambda + 1}{1} = \frac{\mu - \lambda - 2}{2}$,we get $\mu + 2\lambda + 2 = \mu - \lambda - 2 \Rightarrow 3\lambda = -4 \Rightarrow \lambda = -4/3$.
From $\frac{\mu/2 - \lambda - 3}{3} = \frac{\mu/2 + \lambda + 1}{1}$,we get $\mu/2 - \lambda - 3 = 3\mu/2 + 3\lambda + 3 \Rightarrow -\mu = 4\lambda + 6$. Substituting $\lambda = -4/3$,we get $-\mu = 4(-4/3) + 6 = -16/3 + 18/3 = 2/3 \Rightarrow \mu = -2/3$.
Point $M = (2 - 4/3, 4/3, 1 - 4/3) = (2/3, 4/3, -1/3)$.
The equation of line $L$ is $\frac{x-2/3}{3} = \frac{y-4/3}{1} = \frac{z+1/3}{2} = k$.
Any point on $L$ is $(2/3 + 3k, 4/3 + k, -1/3 + 2k)$.
Setting $2/3 + 3k = -1/3 \Rightarrow 3k = -1 \Rightarrow k = -1/3$.
For $k = -1/3$,the point is $(2/3 - 1, 4/3 - 1/3, -1/3 - 2/3) = (-1/3, 1, -1)$.

(A) The given lines are in the form $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$.
From the equations,the points on the lines are $A(3, -7, 1)$ and $B(5, 9, -2)$.
The direction vectors are $\vec{p} = 4\hat{i} - 11\hat{j} + 5\hat{k}$ and $\vec{q} = 3\hat{i} - 6\hat{j} + 1\hat{k}$.
First,find the vector $\vec{AB} = (5-3)\hat{i} + (9-(-7))\hat{j} + (-2-1)\hat{k} = 2\hat{i} + 16\hat{j} - 3\hat{k}$.
Next,find the cross product $\vec{n} = \vec{p} \times \vec{q}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -11 & 5 \\ 3 & -6 & 1 \end{vmatrix} = \hat{i}(-11 + 30) - \hat{j}(4 - 15) + \hat{k}(-24 + 33) = 19\hat{i} + 11\hat{j} + 9\hat{k}$.
The shortest distance ($S$.d.) is the projection of $\vec{AB}$ on $\vec{n}$:
$S.d. = \left| \frac{\vec{AB} \cdot \vec{n}}{|\vec{n}|} \right| = \left| \frac{(2\hat{i} + 16\hat{j} - 3\hat{k}) \cdot (19\hat{i} + 11\hat{j} + 9\hat{k})}{\sqrt{19^2 + 11^2 + 9^2}} \right|$
$S.d. = \left| \frac{38 + 176 - 27}{\sqrt{361 + 121 + 81}} \right| = \frac{187}{\sqrt{563}}$.

(A) The equation of the line $L$ passing through $(1, 2, 3)$ and $(2, 3, 5)$ is given by:
$\frac{x-1}{2-1} = \frac{y-2}{3-2} = \frac{z-3}{5-3} \Rightarrow \frac{x-1}{1} = \frac{y-2}{1} = \frac{z-3}{2} = \lambda$
Any point $B$ on line $L$ is $(1+\lambda, 2+\lambda, 3+2\lambda)$.
The given line along which the distance is measured is $\frac{x-11/3}{2/3} = \frac{y-11/3}{1/3} = \frac{z-19/3}{2/3}$.
This line passes through $A\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$ and has direction ratios $\langle 2, 1, 2 \rangle$.
Since $B$ lies on this line,the vector $\vec{AB}$ must be parallel to $\langle 2, 1, 2 \rangle$.
$\vec{AB} = \left(1+\lambda-\frac{11}{3}, 2+\lambda-\frac{11}{3}, 3+2\lambda-\frac{19}{3}\right) = \left(\lambda-\frac{8}{3}, \lambda-\frac{5}{3}, 2\lambda-\frac{10}{3}\right) = \frac{1}{3} \langle 3\lambda-8, 3\lambda-5, 6\lambda-10 \rangle$.
Since $\vec{AB}$ is parallel to $\langle 2, 1, 2 \rangle$,we have $\frac{3\lambda-8}{2} = \frac{3\lambda-5}{1} = \frac{6\lambda-10}{2}$.
From $\frac{3\lambda-8}{2} = 3\lambda-5$,we get $3\lambda-8 = 6\lambda-10 \Rightarrow 3\lambda = 2 \Rightarrow \lambda = \frac{2}{3}$.
Substituting $\lambda = \frac{2}{3}$ in $\vec{AB}$,we get $\vec{AB} = \frac{1}{3} \langle 3(\frac{2}{3})-8, 3(\frac{2}{3})-5, 6(\frac{2}{3})-10 \rangle = \frac{1}{3} \langle -6, -3, -6 \rangle = \langle -2, -1, -2 \rangle$.
The distance $AB = |\vec{AB}| = \sqrt{(-2)^2 + (-1)^2 + (-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.

(C) Let the given line be $L: \frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4} = \lambda$. Any point $M$ on the line is given by $M(3\lambda+1, 2\lambda, 4\lambda+2)$.
Since $AM$ is perpendicular to the line $L$,the direction vector $\vec{AM} = (3\lambda+1-6, 2\lambda-1, 4\lambda+2-5) = (3\lambda-5, 2\lambda-1, 4\lambda-3)$ must be perpendicular to the direction vector of the line $\vec{b} = (3, 2, 4)$.
Thus,$\vec{AM} \cdot \vec{b} = 0 \implies 3(3\lambda-5) + 2(2\lambda-1) + 4(4\lambda-3) = 0$.
$9\lambda - 15 + 4\lambda - 2 + 16\lambda - 12 = 0 \implies 29\lambda - 29 = 0 \implies \lambda = 1$.
Substituting $\lambda = 1$,we get $M(4, 2, 6)$.
Let $I(x, y, z)$ be the image of $A(6, 1, 5)$ in the line. Since $M$ is the midpoint of $AI$,we have $\frac{x+6}{2} = 4, \frac{y+1}{2} = 2, \frac{z+5}{2} = 6$.
$x+6 = 8 \implies x = 2$; $y+1 = 4 \implies y = 3$; $z+5 = 12 \implies z = 7$.
So,the image point is $I(2, 3, 7)$.
The square of the distance of $I(2, 3, 7)$ from the origin $(0, 0, 0)$ is $2^2 + 3^2 + 7^2 = 4 + 9 + 49 = 62$.

(A) Let the cube $Q$ have vertices $(x_1, x_2, x_3)$ where $x_i \in \{0, 1\}$.
Consider the main diagonal $OG$ connecting $(0,0,0)$ and $(1,1,1)$. Its direction vector is $\vec{v}_1 = (1, 1, 1)$. The equation of line $OG$ is $\frac{x}{1} = \frac{y}{1} = \frac{z}{1}$.
Consider a face diagonal,for example,the diagonal $AB$ on the face $z=0$ connecting $(1,0,0)$ and $(0,1,0)$. Its direction vector is $\vec{v}_2 = (-1, 1, 0)$.
The shortest distance $d$ between two lines with direction vectors $\vec{v}_1, \vec{v}_2$ and points $P_1, P_2$ is given by $d = \frac{|(\vec{P}_2 - \vec{P}_1) \cdot (\vec{v}_1 \times \vec{v}_2)|}{|\vec{v}_1 \times \vec{v}_2|}$.
Here $\vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ -1 & 1 & 0 \end{vmatrix} = \hat{i}(0-1) - \hat{j}(0 - (-1)) + \hat{k}(1 - (-1)) = -\hat{i} - \hat{j} + 2\hat{k}$.
The magnitude is $|\vec{v}_1 \times \vec{v}_2| = \sqrt{(-1)^2 + (-1)^2 + 2^2} = \sqrt{6}$.
Taking $P_1 = (0,0,0)$ and $P_2 = (1,0,0)$,$\vec{P}_2 - \vec{P}_1 = (1,0,0)$.
The distance is $d = \frac{|(1,0,0) \cdot (-1, -1, 2)|}{\sqrt{6}} = \frac{|-1|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$.
For other combinations of face diagonals and main diagonals,the distance is either $0$ (if they intersect) or $\frac{1}{\sqrt{6}}$. Thus,the maximum distance is $\frac{1}{\sqrt{6}}$.

(A) The lines are $L_1: \overrightarrow{r} = (1, 0, 0) + \lambda(-1, 2, 2)$ and $L_2: \overrightarrow{r} = \mu(2, -1, 2)$.
Let $A$ be a point on $L_1$ and $B$ be a point on $L_2$. $A = (1-\lambda, 2\lambda, 2\lambda)$ and $B = (2\mu, -\mu, 2\mu)$.
The vector $\overrightarrow{AB} = (2\mu + \lambda - 1, -\mu - 2\lambda, 2\mu - 2\lambda)$.
The direction of the common perpendicular is $\vec{v} = (-1, 2, 2) \times (2, -1, 2) = (6, 6, -6)$,which is parallel to $(2, 2, -1)$.
Since $AB$ is the shortest distance line,$\overrightarrow{AB}$ must be parallel to $(2, 2, -1)$.
Thus,$\frac{2\mu + \lambda - 1}{2} = \frac{-\mu - 2\lambda}{2} = \frac{2\mu - 2\lambda}{-1} = k$.
Solving this system gives $\lambda = 1/9$ and $\mu = 2/9$.
Then $A = (8/9, 2/9, 2/9)$ and $B = (4/9, -2/9, 4/9)$.
The line $L_3$ passes through $A$ and $B$ with direction $(2, 2, -1)$.
Equation of $L_3$: $\overrightarrow{r} = A + t(2, 2, -1) = (8/9, 2/9, 2/9) + t(2, 2, -1)$.
Option $(1)$ passes through $(2/3, 0, 1/3)$,which is the midpoint of $AB$. Since the direction is $(2, 2, -1)$,it represents $L_3$.
Option $(2)$ passes through $B$,so it represents $L_3$.
Option $(4)$ passes through $(8/9, 2/9, 2/9)$,which is point $A$,so it represents $L_3$.
Thus,options $1, 2, 4$ are correct.

(B) Let the equation of line $l$ be $\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k$. Since $l$ is perpendicular to $l_1$ and $l_2$,its direction ratios $(a, b, c)$ must satisfy the dot product with the direction vectors of $l_1$ and $l_2$ being zero.
Direction of $l_1$ is $\vec{v_1} = (1, 2, 2)$ and direction of $l_2$ is $\vec{v_2} = (2, 2, 1)$.
Thus,$a + 2b + 2c = 0$ and $2a + 2b + c = 0$.
Solving these,we get $\frac{a}{2-4} = \frac{b}{4-1} = \frac{c}{2-4} \Rightarrow \frac{a}{-2} = \frac{b}{3} = \frac{c}{-2}$.
So,line $l$ is $\frac{x}{-2} = \frac{y}{3} = \frac{z}{-2} = k_1$. The point of intersection $P$ of $l$ and $l_1$ is found by setting $(-2k_1, 3k_1, -2k_1) = (3+t, -1+2t, 4+2t)$.
Solving $3+t = -2k_1$ and $-1+2t = 3k_1$ and $4+2t = -2k_1$,we find $k_1 = -1$,so $P = (2, -3, 2)$.
Let a point $Q$ on $l_2$ be $(3+2s, 3+2s, 2+s)$. The distance $PQ = \sqrt{17}$ implies:
$(3+2s-2)^2 + (3+2s+3)^2 + (2+s-2)^2 = 17$
$(1+2s)^2 + (6+2s)^2 + s^2 = 17$
$1 + 4s + 4s^2 + 36 + 24s + 4s^2 + s^2 = 17$
$9s^2 + 28s + 37 = 17 \Rightarrow 9s^2 + 28s + 20 = 0$
$(9s + 10)(s + 2) = 0 \Rightarrow s = -2, s = -\frac{10}{9}$.
For $s = -2$,$Q = (-1, -1, 0)$.
For $s = -\frac{10}{9}$,$Q = (3 - \frac{20}{9}, 3 - \frac{20}{9}, 2 - \frac{10}{9}) = (\frac{7}{9}, \frac{7}{9}, \frac{8}{9})$.
Thus,the points are $(-1, -1, 0)$ and $(\frac{7}{9}, \frac{7}{9}, \frac{8}{9})$,which corresponds to option $(B, D)$.

(D) The lines are given by $L_1: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$ and $L_2: x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha}$.
Rewrite $L_1$ in standard form: $\frac{x-5}{0} = \frac{y}{3-\alpha} = \frac{z}{-2}$.
Rewrite $L_2$ in standard form: $\frac{x-\alpha}{0} = \frac{y}{-1} = \frac{z}{2-\alpha}$.
Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ are coplanar if $\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right| = 0$.
Substituting the values: $\left|\begin{array}{ccc} \alpha-5 & 0 & 0 \\ 0 & 3-\alpha & -2 \\ 0 & -1 & 2-\alpha \end{array}\right| = 0$.
Expanding along the first row: $(\alpha-5) [(3-\alpha)(2-\alpha) - (-2)(-1)] = 0$.
$(\alpha-5) [6 - 3\alpha - 2\alpha + \alpha^2 - 2] = 0$.
$(\alpha-5) (\alpha^2 - 5\alpha + 4) = 0$.
$(\alpha-5)(\alpha-1)(\alpha-4) = 0$.
Thus,$\alpha = 1, 4, 5$. The options provided suggest combinations of these values. Checking the options,$(A, D)$ corresponds to $\alpha = 1, 4$.

(D) Let $P(2\lambda+1, 3\lambda+2, 4\lambda+3)$ be a point on $L_1$ and $Q(3\mu+2, 4\mu+4, 5\mu+5)$ be a point on $L_2$.
The direction ratios of $PQ$ are $(3\mu-2\lambda+1, 4\mu-3\lambda+2, 5\mu-4\lambda+2)$.
Since $PQ$ is the shortest distance line,$PQ \perp L_1$ and $PQ \perp L_2$.
For $PQ \perp L_1$: $2(3\mu-2\lambda+1) + 3(4\mu-3\lambda+2) + 4(5\mu-4\lambda+2) = 0 \Rightarrow 38\mu - 29\lambda + 16 = 0$.
For $PQ \perp L_2$: $3(3\mu-2\lambda+1) + 4(4\mu-3\lambda+2) + 5(5\mu-4\lambda+2) = 0 \Rightarrow 50\mu - 38\lambda + 21 = 0$.
Solving these equations,we get $\lambda = \frac{1}{3}$ and $\mu = -\frac{1}{6}$.
Substituting these values,$P = \left(\frac{5}{3}, 3, \frac{13}{3}\right)$ and $Q = \left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right)$.
The equation of the line $PQ$ passing through $P$ with direction ratios proportional to $\vec{PQ} = Q-P = \left(-\frac{1}{6}, \frac{1}{3}, -\frac{1}{6}\right)$ (or $(1, -2, 1)$) is $\frac{x-5/3}{1} = \frac{y-3}{-2} = \frac{z-13/3}{1}$.
Checking the options,the point $\left(\frac{14}{3}, -3, \frac{22}{3}\right)$ satisfies this equation: $\frac{14/3 - 5/3}{1} = 3$,$\frac{-3-3}{-2} = 3$,$\frac{22/3 - 13/3}{1} = 3$.

(D) For lines $L_1$ and $L_2$ to intersect at point $B$,the coordinates must satisfy:
$(3\lambda+1, -\lambda+1, -1) = (2\mu+2, 0, \alpha\mu-4)$.
From the $y$-coordinate,$-\lambda+1 = 0 \implies \lambda = 1$.
Substituting $\lambda=1$ into the $x$-coordinate: $3(1)+1 = 2\mu+2 \implies 4 = 2\mu+2 \implies \mu = 1$.
Substituting $\mu=1$ into the $z$-coordinate: $-1 = \alpha(1)-4 \implies \alpha = 3$.
Thus,point $B = (4, 0, -1)$.
Line $L_2$ is given by $\frac{x-2}{2} = \frac{y}{0} = \frac{z+4}{3} = \delta$.
So,any point $P$ on $L_2$ is $(2\delta+2, 0, 3\delta-4)$.
The direction vector of $AP$ is $\vec{AP} = (2\delta+2-1, 0-1, 3\delta-4+1) = (2\delta+1, -1, 3\delta-3)$.
Since $AP \perp L_2$,the dot product of $\vec{AP}$ and the direction vector of $L_2$ $(2, 0, 3)$ is zero:
$2(2\delta+1) + 0(-1) + 3(3\delta-3) = 0
\implies 4\delta+2 + 9\delta-9 = 0
\implies 13\delta = 7 \implies \delta = \frac{7}{13}$.
Point $P = (2(\frac{7}{13})+2, 0, 3(\frac{7}{13})-4) = (\frac{40}{13}, 0, -\frac{31}{13})$.
$PB^2 = (4-\frac{40}{13})^2 + (0-0)^2 + (-1+\frac{31}{13})^2 = (\frac{12}{13})^2 + (\frac{18}{13})^2 = \frac{144+324}{169} = \frac{468}{169}$.
Finally,$26\alpha(PB)^2 = 26 \times 3 \times \frac{468}{169} = 78 \times \frac{36}{13} = 6 \times 36 = 216$.

(A) The line passes through $P(-2, -1, 3)$ and is parallel to $\vec{v} = 3\hat{i} + 2\hat{j} + 2\hat{k}$. Thus,the coordinates of $Q$ can be written as $Q(3\lambda - 2, 2\lambda - 1, 2\lambda + 3)$ for some $\lambda \neq 0$.
Given the distance $QR = 5$,we have $\sqrt{(3\lambda - 2 - 1)^2 + (2\lambda - 1 - 3)^2 + (2\lambda + 3 - 3)^2} = 5$.
Squaring both sides: $(3\lambda - 3)^2 + (2\lambda - 4)^2 + (2\lambda)^2 = 25$.
$9(\lambda - 1)^2 + 4(\lambda - 2)^2 + 4\lambda^2 = 25$.
$9(\lambda^2 - 2\lambda + 1) + 4(\lambda^2 - 4\lambda + 4) + 4\lambda^2 = 25$.
$17\lambda^2 - 34\lambda + 25 = 25 \Rightarrow 17\lambda(\lambda - 2) = 0$.
Since $Q$ is distinct from $P$,$\lambda \neq 0$,so $\lambda = 2$.
Thus,$Q = (3(2) - 2, 2(2) - 1, 2(2) + 3) = (4, 3, 7)$.
Now,$\vec{PQ} = Q - P = (4 - (-2), 3 - (-1), 7 - 3) = (6, 4, 4)$.
$\vec{PR} = R - P = (1 - (-2), 3 - (-1), 3 - 3) = (3, 4, 0)$.
The area of $\triangle PQR$ is $\frac{1}{2} |\vec{PQ} \times \vec{PR}|$.
$\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 4 & 4 \\ 3 & 4 & 0 \end{vmatrix} = \hat{i}(0 - 16) - \hat{j}(0 - 12) + \hat{k}(24 - 12) = -16\hat{i} + 12\hat{j} + 12\hat{k}$.
$|\vec{PQ} \times \vec{PR}| = \sqrt{(-16)^2 + 12^2 + 12^2} = \sqrt{256 + 144 + 144} = \sqrt{544}$.
Area $= \frac{1}{2} \sqrt{544} = \sqrt{\frac{544}{4}} = \sqrt{136}$.
Therefore,the square of the area is $136$.

(C) Let the given line be $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}=\lambda$.
Any point $A$ on the line is given by $(2\lambda+1, -\lambda-2, 2\lambda-3)$.
The vector $\vec{PA} = (2\lambda+1-2, -\lambda-2-(-10), 2\lambda-3-1) = (2\lambda-1, -\lambda+8, 2\lambda-4)$.
The direction vector of the line is $\vec{n} = 2\hat{i} - \hat{j} + 2\hat{k}$.
Since $PA$ is perpendicular to the line,$\vec{PA} \cdot \vec{n} = 0$.
$(2\lambda-1)(2) + (-\lambda+8)(-1) + (2\lambda-4)(2) = 0$.
$4\lambda - 2 + \lambda - 8 + 4\lambda - 8 = 0$.
$9\lambda - 18 = 0 \Rightarrow \lambda = 2$.
Substituting $\lambda = 2$ in the coordinates of $A$,we get $A(2(2)+1, -2-2, 2(2)-3) = A(5, -4, 1)$.
The perpendicular distance $AP$ is the distance between $P(2, -10, 1)$ and $A(5, -4, 1)$.
$AP = \sqrt{(5-2)^2 + (-4 - (-10))^2 + (1-1)^2} = \sqrt{3^2 + 6^2 + 0^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$.

(D) Let the line be $L: \frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2} = \lambda$. Any point on the line is $P(7\lambda+3, -\lambda+2, -2\lambda-1)$.
Since $P$ is the foot of the perpendicular from $Q(10,-3,-1)$ to the line,the vector $\vec{QP}$ must be perpendicular to the direction vector of the line $\vec{v} = 7\hat{i} - \hat{j} - 2\hat{k}$.
$\vec{QP} = (7\lambda+3-10)\hat{i} + (-\lambda+2+3)\hat{j} + (-2\lambda-1+1)\hat{k} = (7\lambda-7)\hat{i} + (-\lambda+5)\hat{j} - 2\lambda\hat{k}$.
Since $\vec{QP} \cdot \vec{v} = 0$,we have $7(7\lambda-7) - 1(-\lambda+5) - 2(-2\lambda) = 0$.
$49\lambda - 49 + \lambda - 5 + 4\lambda = 0 \Rightarrow 54\lambda - 54 = 0 \Rightarrow \lambda = 1$.
Thus,$P = (7(1)+3, -1+2, -2(1)-1) = (10, 1, -3)$.
Now,$\vec{PQ} = (10-10)\hat{i} + (-3-1)\hat{j} + (-1-(-3))\hat{k} = -4\hat{j} + 2\hat{k}$.
And $\vec{PR} = (3-10)\hat{i} + (-2-1)\hat{j} + (1-(-3))\hat{k} = -7\hat{i} - 3\hat{j} + 4\hat{k}$.
The area of $\triangle PQR = \frac{1}{2} |\vec{PQ} \times \vec{PR}|$.
$\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -4 & 2 \\ -7 & -3 & 4 \end{vmatrix} = \hat{i}(-16 - (-6)) - \hat{j}(0 - (-14)) + \hat{k}(0 - 28) = -10\hat{i} - 14\hat{j} - 28\hat{k}$.
Magnitude $|\vec{PQ} \times \vec{PR}| = \sqrt{(-10)^2 + (-14)^2 + (-28)^2} = \sqrt{100 + 196 + 784} = \sqrt{1080} = \sqrt{36 \times 30} = 6\sqrt{30}$.
Area $= \frac{1}{2} \times 6\sqrt{30} = 3\sqrt{30}$.

(B) The lines are given by $\vec{r} = \vec{a_1} + \lambda \vec{p}$ and $\vec{r} = \vec{a_2} + \mu \vec{q}$,where $\vec{a_1} = (2, 1, -3)$,$\vec{p} = (1, 2, -3)$,$\vec{a_2} = (-1, -3, -5)$,and $\vec{q} = (2, 4, -5)$.
The shortest distance $d$ between two skew lines is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|}$.
First,calculate $\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10 + 12) - \hat{j}(-5 + 6) + \hat{k}(4 - 4) = 2\hat{i} - \hat{j} + 0\hat{k}$.
The magnitude is $|\vec{p} \times \vec{q}| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{5}$.
Next,$\vec{a_2} - \vec{a_1} = (-1-2, -3-1, -5-(-3)) = (-3, -4, -2)$.
The dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} \times \vec{q}) = (-3, -4, -2) \cdot (2, -1, 0) = -6 + 4 + 0 = -2$.
Thus,$d = \frac{|-2|}{\sqrt{5}} = \frac{2}{\sqrt{5}}$.
The square of the shortest distance is $d^2 = \frac{4}{5}$.
Given $d^2 = \frac{m}{n} = \frac{4}{5}$,where $m=4$ and $n=5$ are coprime.
Therefore,$m+n = 4+5 = 9$.

(B) Let $M$ be the foot of the perpendicular from $B(1,2,3)$ to the line $L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2} = \lambda$.
Any point on the line is $M(3\lambda+6, 2\lambda+7, -2\lambda+7)$.
The vector $\overrightarrow{BM} = (3\lambda+6-1)\hat{i} + (2\lambda+7-2)\hat{j} + (-2\lambda+7-3)\hat{k} = (3\lambda+5)\hat{i} + (2\lambda+5)\hat{j} + (-2\lambda+4)\hat{k}$.
Since $\overrightarrow{BM}$ is perpendicular to the direction vector of the line $\vec{v} = 3\hat{i} + 2\hat{j} - 2\hat{k}$,we have $\overrightarrow{BM} \cdot \vec{v} = 0$.
$3(3\lambda+5) + 2(2\lambda+5) - 2(-2\lambda+4) = 0$.
$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0 \implies 17\lambda + 17 = 0 \implies \lambda = -1$.
Substituting $\lambda = -1$,we get $\overrightarrow{BM} = (3(-1)+5)\hat{i} + (2(-1)+5)\hat{j} + (-2(-1)+4)\hat{k} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.
The length of the altitude $BM = |\overrightarrow{BM}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$.
The area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times BM = \frac{1}{2} \times 6 \times 7 = 21$ sq. units.

(D) The equation of the line passing through $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ is given by:
$\frac{x+1}{2}=\frac{y-2}{3}=\frac{z-1}{4}=\lambda$
So,the coordinates of any point on this line are $(2\lambda-1, 3\lambda+2, 4\lambda+1)$.
The equation of the second line is given by:
$\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}=\mu$
So,the coordinates of any point on this line are $(3\mu-2, 2\mu+3, \mu+4)$.
For the intersection point $P$,the coordinates must be equal:
$2\lambda-1 = 3\mu-2 \implies 2\lambda - 3\mu = -1$ $(i)$
$3\lambda+2 = 2\mu+3 \implies 3\lambda - 2\mu = 1$ (ii)
Solving $(i)$ and (ii),we multiply $(i)$ by $2$ and (ii) by $3$:
$4\lambda - 6\mu = -2$
$9\lambda - 6\mu = 3$
Subtracting these gives $5\lambda = 5$,so $\lambda = 1$.
Substituting $\lambda = 1$ into $(i)$,$2(1) - 3\mu = -1 \implies 3\mu = 3 \implies \mu = 1$.
Checking with the $z$-coordinate: $4(1)+1 = 5$ and $1+4 = 5$. Since they match,the intersection point $P$ is $(2(1)-1, 3(1)+2, 4(1)+1) = (1, 5, 5)$.
The distance of $P(1, 5, 5)$ from $Q(4, -5, 1)$ is:
$PQ = \sqrt{(4-1)^2 + (-5-5)^2 + (1-5)^2}$
$PQ = \sqrt{3^2 + (-10)^2 + (-4)^2}$
$PQ = \sqrt{9 + 100 + 16} = \sqrt{125} = 5\sqrt{5}$.

(B) Let the line $L$ be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4} = \lambda$. Any point on $L$ is $(2\lambda+1, 3\lambda-1, 4\lambda)$.
Since $R(5, p, q)$ lies on $L$,we have $2\lambda+1 = 5 \implies \lambda = 2$. Thus,$R = (5, 5, 8)$.
Let $T$ be the foot of the perpendicular from $Q(7, -2, 5)$ to the line $L$. Let $T = (2\lambda+1, 3\lambda-1, 4\lambda)$.
The vector $\vec{QT} = (2\lambda+1-7, 3\lambda-1+2, 4\lambda-5) = (2\lambda-6, 3\lambda+1, 4\lambda-5)$.
Since $\vec{QT}$ is perpendicular to the direction vector of the line $\vec{b} = (2, 3, 4)$,we have $\vec{QT} \cdot \vec{b} = 0$.
$2(2\lambda-6) + 3(3\lambda+1) + 4(4\lambda-5) = 0 \implies 4\lambda - 12 + 9\lambda + 3 + 16\lambda - 20 = 0 \implies 29\lambda = 29 \implies \lambda = 1$.
Thus,$T = (3, 2, 4)$.
$P$ is the image of $Q$ in $L$,so $T$ is the midpoint of $PQ$. Thus,$QT = TP$.
The area of $\triangle PQR = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times RT \times (QT + TP) = \frac{1}{2} \times RT \times (2QT) = RT \times QT$.
$QT = \sqrt{(3-7)^2 + (2+2)^2 + (4-5)^2} = \sqrt{(-4)^2 + 4^2 + (-1)^2} = \sqrt{16+16+1} = \sqrt{33}$.
$RT = \sqrt{(5-3)^2 + (5-2)^2 + (8-4)^2} = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4+9+16} = \sqrt{29}$.
Area of $\triangle PQR = \sqrt{33} \times \sqrt{29} = \sqrt{957}$.
Therefore,the square of the area is $(\sqrt{957})^2 = 957$.

(A) Let the given line be $L: \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3} = \lambda$.
Any point on the line is $Q(2\lambda+1, \lambda+2, 3\lambda+1)$.
Let $P = (4,4,3)$. The vector $\vec{PQ} = (2\lambda+1-4, \lambda+2-4, 3\lambda+1-3) = (2\lambda-3, \lambda-2, 3\lambda-2)$.
Since $\vec{PQ}$ is perpendicular to the line with direction vector $\vec{v} = (2, 1, 3)$,we have $\vec{PQ} \cdot \vec{v} = 0$.
$2(2\lambda-3) + 1(\lambda-2) + 3(3\lambda-2) = 0$.
$4\lambda - 6 + \lambda - 2 + 9\lambda - 6 = 0$.
$14\lambda - 14 = 0 \Rightarrow \lambda = 1$.
Thus,the foot of the perpendicular $Q$ is $(2(1)+1, 1+2, 3(1)+1) = (3, 3, 4)$.
Let the image of $P$ be $R(\alpha, \beta, \gamma)$. Since $Q$ is the midpoint of $PR$,we have:
$\frac{\alpha+4}{2} = 3 \Rightarrow \alpha = 2$.
$\frac{\beta+4}{2} = 3 \Rightarrow \beta = 2$.
$\frac{\gamma+3}{2} = 4 \Rightarrow \gamma = 5$.
So,the image is $(2, 2, 5)$.
Therefore,$\alpha+\beta+\gamma = 2+2+5 = 9$.

(C) Let the given point be $P\left(\frac{15}{7}, \frac{32}{7}, 7\right)$ and the line $L$ be $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} = k$.
Any point $Q$ on the line $L$ is given by $Q(3k-1, 5k-3, 7k-5)$.
The vector $\vec{PQ}$ is given by $\vec{PQ} = \left(3k-1-\frac{15}{7}\right)\hat{i} + \left(5k-3-\frac{32}{7}\right)\hat{j} + (7k-5-7)\hat{k} = \left(3k-\frac{22}{7}\right)\hat{i} + \left(5k-\frac{53}{7}\right)\hat{j} + (7k-12)\hat{k}$.
Since the line $PQ$ is parallel to the vector $\vec{v} = \hat{i} + 4\hat{j} + 7\hat{k}$,the components of $\vec{PQ}$ must be proportional to the components of $\vec{v}$:
$\frac{3k-\frac{22}{7}}{1} = \frac{5k-\frac{53}{7}}{4} = \frac{7k-12}{7} = \lambda$.
From $\frac{7k-12}{7} = \lambda$,we have $7k-12 = 7\lambda \Rightarrow k = \lambda + \frac{12}{7}$.
Substituting $k$ into the first equality: $3(\lambda + \frac{12}{7}) - \frac{22}{7} = \lambda \Rightarrow 3\lambda + \frac{36-22}{7} = \lambda \Rightarrow 2\lambda = -2 \Rightarrow \lambda = -1$.
Thus,$k = -1 + \frac{12}{7} = \frac{5}{7}$.
The point $Q$ is $Q(3(\frac{5}{7})-1, 5(\frac{5}{7})-3, 7(\frac{5}{7})-5) = Q(\frac{8}{7}, \frac{4}{7}, 0)$.
The distance $PQ = \sqrt{(\frac{15}{7}-\frac{8}{7})^2 + (\frac{32}{7}-\frac{4}{7})^2 + (7-0)^2} = \sqrt{1^2 + 4^2 + 7^2} = \sqrt{1+16+49} = \sqrt{66}$.
Therefore,the square of the distance is $(PQ)^2 = 66$.

(B) The direction vector of line $L$ is perpendicular to the direction vectors of the two given lines,$\vec{v_1} = (2, 1, -2)$ and $\vec{v_2} = (1, 3, 4)$.
Thus,the direction vector $\vec{v}$ of $L$ is $\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix} = \hat{i}(4 - (-6)) - \hat{j}(8 - (-2)) + \hat{k}(6 - 1) = 10\hat{i} - 10\hat{j} + 5\hat{k}$.
We can take the direction vector as $\vec{d} = (2, -2, 1)$.
The equation of line $L$ passing through $P(2, -1, 3)$ is $\frac{x-2}{2} = \frac{y+1}{-2} = \frac{z-3}{1} = \lambda$.
Any point $Q$ on line $L$ is of the form $(2\lambda + 2, -2\lambda - 1, \lambda + 3)$.
Since $Q$ lies on the $yz$-plane,its $x$-coordinate must be $0$.
$2\lambda + 2 = 0 \implies \lambda = -1$.
Substituting $\lambda = -1$ into the coordinates of $Q$,we get $Q(0, -2(-1) - 1, -1 + 3) = Q(0, 1, 2)$.
The distance $PQ = \sqrt{(2-0)^2 + (-1-1)^2 + (3-2)^2} = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.

(B) The line $L$ is given by $\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z-2}{2} = \mu$. Any point on $L$ is $P(\mu+1, -\mu-1, 2\mu+2)$.
Since $AP \perp L$,the vector $\overrightarrow{AP} = (\mu+1-1, -\mu-1-2, 2\mu+2-2) = (\mu, -\mu-3, 2\mu)$ is perpendicular to the direction vector of $L$,$\vec{d} = (1, -1, 2)$.
$\overrightarrow{AP} \cdot \vec{d} = 0 \Rightarrow (\mu)(1) + (-\mu-3)(-1) + (2\mu)(2) = 0$.
$\mu + \mu + 3 + 4\mu = 0 \Rightarrow 6\mu = -3 \Rightarrow \mu = -\frac{1}{2}$.
Thus,$P = (-\frac{1}{2}+1, -(-\frac{1}{2})-1, 2(-\frac{1}{2})+2) = (\frac{1}{2}, -\frac{1}{2}, 1)$.
The line $L_2$ is given by $\vec{r} = (-1, 1, -2) + \lambda(1, -1, 1)$. Any point on $L_2$ is $Q(-1+\lambda, 1-\lambda, -2+\lambda)$.
Since $Q$ lies on $L$,it must satisfy the equation of $L$: $\frac{(-1+\lambda)-1}{1} = \frac{(1-\lambda)+1}{-1} = \frac{(-2+\lambda)-2}{2} = \mu$.
$\frac{\lambda-2}{1} = \frac{2-\lambda}{-1} = \frac{\lambda-4}{2}$.
From $\lambda-2 = \lambda-2$ (always true) and $\lambda-2 = \frac{\lambda-4}{2} \Rightarrow 2\lambda-4 = \lambda-4 \Rightarrow \lambda = 0$.
For $\lambda = 0$,$Q = (-1, 1, -2)$.
Now,$PQ^2 = (\frac{1}{2} - (-1))^2 + (-\frac{1}{2} - 1)^2 + (1 - (-2))^2 = (\frac{3}{2})^2 + (-\frac{3}{2})^2 + (3)^2 = \frac{9}{4} + \frac{9}{4} + 9 = \frac{18}{4} + 9 = 4.5 + 9 = 13.5$.
Therefore,$2(PQ)^2 = 2(13.5) = 27$.

(B) Let $M$ be the foot of the perpendicular from $P(0,2,3)$ to the line $L: \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3} = \lambda$.
Any point on the line is $M(5\lambda-3, 2\lambda+1, 3\lambda-4)$.
The direction ratios of $PM$ are $(5\lambda-3-0, 2\lambda+1-2, 3\lambda-4-3) = (5\lambda-3, 2\lambda-1, 3\lambda-7)$.
Since $PM \perp L$,the dot product of the direction ratios of $PM$ and $L$ is zero:
$5(5\lambda-3) + 2(2\lambda-1) + 3(3\lambda-7) = 0$
$25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0$
$38\lambda - 38 = 0 \Rightarrow \lambda = 1$.
Thus,$M = (5(1)-3, 2(1)+1, 3(1)-4) = (2, 3, -1)$.
The length of the altitude $PM$ is $\sqrt{(2-0)^2 + (3-2)^2 + (-1-3)^2} = \sqrt{4+1+16} = \sqrt{21}$.
The area of $\triangle PQR = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times QR \times PM = \frac{1}{2} \times 5 \times \sqrt{21} = \frac{5\sqrt{21}}{2}$.
Given area $= \frac{m}{n} = \frac{5\sqrt{21}}{2}$.
Comparing,we get $2m = 5\sqrt{21}n$,which implies $2m - 5\sqrt{21}n = 0$.

(B) The line passing through $A(4, 7, 1)$ and $B(3, 5, 3)$ has direction ratios $(3-4, 5-7, 3-1) = (-1, -2, 2)$.
The equation of the line $AB$ is $\frac{x-3}{-1} = \frac{y-5}{-2} = \frac{z-3}{2} = \lambda$.
Any point $R$ on the line is $(\lambda+3, 2\lambda+5, -2\lambda+3)$.
The vector $\vec{PR} = (\lambda+3-1, 2\lambda+5-0, -2\lambda+3-3) = (\lambda+2, 2\lambda+5, -2\lambda)$.
Since $PR \perp AB$,the dot product of $\vec{PR}$ and the direction vector of $AB$ $(-1, -2, 2)$ is $0$:
$-1(\lambda+2) - 2(2\lambda+5) + 2(-2\lambda) = 0$.
$-\lambda - 2 - 4\lambda - 10 - 4\lambda = 0 \Rightarrow -9\lambda = 12 \Rightarrow \lambda = -\frac{4}{3}$.
The foot of the perpendicular $R$ is $(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$.
Since $R$ is the midpoint of $PQ$,$Q = 2R - P = (2 \times \frac{5}{3} - 1, 2 \times \frac{7}{3} - 0, 2 \times \frac{17}{3} - 3) = (\frac{7}{3}, \frac{14}{3}, \frac{25}{3})$.
Thus,$\alpha + \beta + \gamma = \frac{7+14+25}{3} = \frac{46}{3}$.

(A) The equations of the lines are $L_1: \vec{r} = (7 \hat{i} + 6 \hat{j} + 2 \hat{k}) + \lambda(-3 \hat{i} + 2 \hat{j} + 4 \hat{k})$ and $L_2: \vec{r} = (5 \hat{i} + 3 \hat{j} + 4 \hat{k}) + \mu(2 \hat{i} + \hat{j} + 3 \hat{k})$.
Let $\vec{a_1} = 7 \hat{i} + 6 \hat{j} + 2 \hat{k}$,$\vec{a_2} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}$,$\vec{v_1} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$,and $\vec{v_2} = 2 \hat{i} + \hat{j} + 3 \hat{k}$.
The shortest distance $d$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{v_1} \times \vec{v_2})|}{|\vec{v_1} \times \vec{v_2}|}$.
First,calculate $\vec{a_2} - \vec{a_1} = (5-7)\hat{i} + (3-6)\hat{j} + (4-2)\hat{k} = -2 \hat{i} - 3 \hat{j} + 2 \hat{k}$.
Next,calculate $\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 4 \\ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-4) - \hat{j}(-9-8) + \hat{k}(-3-4) = 2 \hat{i} + 17 \hat{j} - 7 \hat{k}$.
The magnitude $|\vec{v_1} \times \vec{v_2}| = \sqrt{2^2 + 17^2 + (-7)^2} = \sqrt{4 + 289 + 49} = \sqrt{342} = \sqrt{9 \times 38} = 3 \sqrt{38}$.
The dot product $|(\vec{a_2} - \vec{a_1}) \cdot (\vec{v_1} \times \vec{v_2})| = |(-2)(2) + (-3)(17) + (2)(-7)| = |-4 - 51 - 14| = |-69| = 69$.
Thus,$d = \frac{69}{3 \sqrt{38}} = \frac{23}{\sqrt{38}}$.

(A) Let the line $L_1$ be $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=p$. Then $A = (2p+1, 3p+2, 4p+3)$.
Let the line $L_2$ be $\frac{x-6}{1}=\frac{y}{1}=\frac{z-4}{-1}=q$. Then $B = (q+6, q, 4-q)$.
The points $P(4,1,0)$,$A$,and $B$ are collinear. The direction ratios of $PA$ are $(2p+1-4, 3p+2-1, 4p+3-0) = (2p-3, 3p+1, 4p+3)$.
The direction ratios of $AB$ are $(q+6-(2p+1), q-(3p+2), 4-q-(4p+3)) = (q-2p+5, q-3p-2, -q-4p+1)$.
Since $P, A, B$ are collinear,the direction ratios of $PA$ and $AB$ are proportional:
$\frac{2p-3}{q-2p+5} = \frac{3p+1}{q-3p-2} = \frac{4p+3}{-q-4p+1} = k$.
Solving this system,we find $p=-1$ and $q=3$.
Substituting $p=-1$ into $A$,we get $A(-1, -1, -1)$.
Substituting $q=3$ into $B$,we get $B(9, 3, 1)$.
Now,calculate the determinant:
$\left|\begin{array}{ccc}1 & 0 & 1 \\ -1 & -1 & -1 \\ 9 & 3 & 1\end{array}\right| = 1(-1 - (-3)) - 0 + 1(-3 - (-9)) = 1(2) + 1(6) = 8$.

(C) The equation of line $L_1$ passing through $(1, 2, 3)$ and parallel to the $z$-axis is $\frac{x-1}{0} = \frac{y-2}{0} = \frac{z-3}{1}$.
The equation of line $L_2$ passing through $(\lambda, 5, 6)$ and parallel to the $y$-axis is $\frac{x-\lambda}{0} = \frac{y-5}{1} = \frac{z-6}{0}$.
The shortest distance $SD$ between two lines $\vec{r} = \vec{a_1} + t\vec{b_1}$ and $\vec{r} = \vec{a_2} + s\vec{b_2}$ is given by $\frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$.
Here,$\vec{a_1} = (1, 2, 3)$,$\vec{b_1} = (0, 0, 1)$,$\vec{a_2} = (\lambda, 5, 6)$,$\vec{b_2} = (0, 1, 0)$.
$\vec{a_2} - \vec{a_1} = (\lambda-1, 3, 3)$.
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{vmatrix} = -\hat{i}$.
$SD = \frac{|(\lambda-1, 3, 3) \cdot (-1, 0, 0)|}{|-1|} = |-(\lambda-1)| = |\lambda-1| = 3$.
Thus,$\lambda-1 = 3$ or $\lambda-1 = -3$,so $\lambda = 4$ or $\lambda = -2$.
Given $\lambda_2 < \lambda_1$,we have $\lambda_1 = 4$ and $\lambda_2 = -2$.
The point is $P(4, -2, 7)$. Line $L_1$ is $(1, 2, z)$.
The distance squared from $P(4, -2, 7)$ to $L_1$ is the distance to the point $(1, 2, 7)$ on $L_1$ (since the $z$-coordinate of $P$ is $7$,the projection onto $L_1$ is $(1, 2, 7)$).
$PQ^2 = (4-1)^2 + (-2-2)^2 + (7-7)^2 = 3^2 + (-4)^2 + 0^2 = 9 + 16 = 25$.

(B) Let the given line be $L_1: \frac{x-4}{1} = \frac{y-4}{0} = \frac{z-2}{3} = \lambda$. Any point $Q$ on this line is $(\lambda+4, 4, 3\lambda+2)$.
Since the distance is measured along the line $\frac{x-9}{2} = \frac{y-13}{3} = \frac{z-17}{6}$,the vector $\vec{PQ}$ must be parallel to the vector $\vec{v} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.
Vector $\vec{PQ} = Q - P = (\lambda+4-7, 4-10, 3\lambda+2-11) = (\lambda-3, -6, 3\lambda-9)$.
Since $\vec{PQ}$ is parallel to $\vec{v}$,their components must be proportional:
$\frac{\lambda-3}{2} = \frac{-6}{3} = \frac{3\lambda-9}{6}$.
From $\frac{\lambda-3}{2} = -2$,we get $\lambda-3 = -4$,so $\lambda = -1$.
Substituting $\lambda = -1$ into the coordinates of $Q$,we get $Q = (-1+4, 4, 3(-1)+2) = (3, 4, -1)$.
The distance $PQ = \sqrt{(3-7)^2 + (4-10)^2 + (-1-11)^2} = \sqrt{(-4)^2 + (-6)^2 + (-12)^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$.

(B) The lines are $L_1: \frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $L_2: \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$.
Points on the lines are $A(3, \alpha, 3)$ and $B(-3, -7, \beta)$.
Direction vectors are $\vec{p} = (3, -1, 1)$ and $\vec{q} = (-3, 2, 4)$.
Vector $\vec{BA} = (3 - (-3))\hat{i} + (\alpha - (-7))\hat{j} + (3 - \beta)\hat{k} = 6\hat{i} + (\alpha+7)\hat{j} + (3-\beta)\hat{k}$.
Cross product $\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} = \hat{i}(-4-2) - \hat{j}(12 - (-3)) + \hat{k}(6-3) = -6\hat{i} - 15\hat{j} + 3\hat{k}$.
The magnitude $|\vec{p} \times \vec{q}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}$.
Shortest distance $d = \frac{|\vec{BA} \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|} = 3\sqrt{30}$.
$|6(-6) + (\alpha+7)(-15) + (3-\beta)(3)| = (3\sqrt{30})^2 = 270$.
$|-36 - 15\alpha - 105 + 9 - 3\beta| = 270$.
$|-15\alpha - 3\beta - 132| = 270$.
Since we need a positive value,we take $15\alpha + 3\beta + 132 = 270$ or $15\alpha + 3\beta + 132 = -270$.
$15\alpha + 3\beta = 138 \implies 5\alpha + \beta = 46$.

(B) Let the line $L$ be $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} = \lambda$. Any point on the line is $Q(3\lambda+6, 2\lambda+7, -2\lambda+7)$.
Let $P = (1, 2, 3)$. The vector $\overrightarrow{PQ} = (3\lambda+5, 2\lambda+5, -2\lambda+4)$.
Since $PQ$ is perpendicular to the line $L$ with direction vector $\vec{b} = (3, 2, -2)$,we have $\overrightarrow{PQ} \cdot \vec{b} = 0$.
$3(3\lambda+5) + 2(2\lambda+5) - 2(-2\lambda+4) = 0$
$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0$
$17\lambda + 17 = 0 \Rightarrow \lambda = -1$.
Thus,the foot of the perpendicular is $Q(3(-1)+6, 2(-1)+7, -2(-1)+7) = Q(3, 5, 9)$.
Points $A$ and $B$ are at a distance $d = 2\sqrt{17}$ from $Q$ on the line. Let the unit vector along the line be $\hat{u} = \frac{1}{\sqrt{3^2+2^2+(-2)^2}}(3, 2, -2) = \frac{1}{\sqrt{17}}(3, 2, -2)$.
The points $A$ and $B$ are given by $Q \pm d\hat{u} = (3, 5, 9) \pm 2\sqrt{17} \cdot \frac{1}{\sqrt{17}}(3, 2, -2) = (3, 5, 9) \pm 2(3, 2, -2)$.
$A = (3+6, 5+4, 9-4) = (9, 9, 5)$ and $B = (3-6, 5-4, 9+4) = (-3, 1, 13)$.
Then $\overrightarrow{OA} \cdot \overrightarrow{OB} = (9, 9, 5) \cdot (-3, 1, 13) = (9)(-3) + (9)(1) + (5)(13) = -27 + 9 + 65 = 47$.

(C) The shortest distance between two lines $\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{p}$ and $\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{q}$ is given by $d = \frac{|(\overrightarrow{a_2} - \overrightarrow{a_1}) \cdot (\overrightarrow{p} \times \overrightarrow{q})|}{|\overrightarrow{p} \times \overrightarrow{q}|}$.
Here,$\overrightarrow{a_1} = -\hat{i}$,$\overrightarrow{p} = 3\hat{i} + 4\hat{j} + 5\hat{k}$,$\overrightarrow{a_2} = p\hat{i} + 2\hat{j} + \hat{k}$,and $\overrightarrow{q} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
$\overrightarrow{a_2} - \overrightarrow{a_1} = (p+1)\hat{i} + 2\hat{j} + \hat{k}$.
$\overrightarrow{p} \times \overrightarrow{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{vmatrix} = \hat{i}(16-15) - \hat{j}(12-10) + \hat{k}(9-8) = \hat{i} - 2\hat{j} + \hat{k}$.
$|\overrightarrow{p} \times \overrightarrow{q}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$.
Shortest distance $d = \frac{|((p+1)\hat{i} + 2\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k})|}{\sqrt{6}} = \frac{|p+1 - 4 + 1|}{\sqrt{6}} = \frac{|p-2|}{\sqrt{6}}$.
Given $d = \frac{1}{\sqrt{6}}$,so $|p-2| = 1$,which implies $p-2 = 1$ or $p-2 = -1$.
Thus,$p = 3$ or $p = 1$. Since $a < b$,we have $a = 1$ and $b = 3$.
The ellipse is $\frac{x^2}{1^2} + \frac{y^2}{3^2} = 1$. Here $a^2 = 1$ and $b^2 = 9$,so $b > a$.
The length of the latus rectum for $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $b > a$ is $\frac{2a^2}{b} = \frac{2(1)}{3} = \frac{2}{3}$.

(A) The direction vectors of the lines $L_1$ and $L_2$ are $\vec{v_1} = \langle 1, 0, -1 \rangle$ and $\vec{v_2} = \langle 3, 4, 5 \rangle$ respectively.
Let $\theta$ be the angle between the lines $L_1$ and $L_2$.
$\cos \theta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|} = \frac{|(1)(3) + (0)(4) + (-1)(5)|}{\sqrt{1^2 + 0^2 + (-1)^2} \sqrt{3^2 + 4^2 + 5^2}} = \frac{|3 + 0 - 5|}{\sqrt{2} \sqrt{9 + 16 + 25}} = \frac{2}{\sqrt{2} \sqrt{50}} = \frac{2}{\sqrt{100}} = \frac{2}{10} = \frac{1}{5}$.
Since $\cos \theta = \frac{1}{5}$,we have $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (\frac{1}{5})^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5}$.
The area of triangle $ABC$ is given by $\text{Area} = \frac{1}{2} \times AB \times AC \times \sin \theta$.
Given $AB = AC = \sqrt{15}$,we have $\text{Area} = \frac{1}{2} \times \sqrt{15} \times \sqrt{15} \times \frac{\sqrt{24}}{5} = \frac{1}{2} \times 15 \times \frac{\sqrt{24}}{5} = \frac{3 \sqrt{24}}{2}$.
The square of the area is $(\frac{3 \sqrt{24}}{2})^2 = \frac{9 \times 24}{4} = 9 \times 6 = 54$.

(D) Let the two lines be $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$.
Points on the lines are $A(1, 2, 3)$ and $B(0, 0, 5)$.
Direction vectors are $\vec{b_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b_2} = \hat{i} + \alpha\hat{j} + \hat{k}$.
The cross product $\vec{n} = \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & \alpha & 1 \end{vmatrix} = \hat{i}(3-4\alpha) - \hat{j}(2-4) + \hat{k}(2\alpha-3) = (3-4\alpha)\hat{i} + 2\hat{j} + (2\alpha-3)\hat{k}$.
Vector $\vec{AB} = (0-1)\hat{i} + (0-2)\hat{j} + (5-3)\hat{k} = -\hat{i} - 2\hat{j} + 2\hat{k}$.
Shortest distance $d = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{n}|} = \frac{|(-1)(3-4\alpha) + (-2)(2) + (2)(2\alpha-3)|}{\sqrt{(3-4\alpha)^2 + 2^2 + (2\alpha-3)^2}} = \frac{5}{\sqrt{6}}$.
$|4\alpha - 3 - 4 + 4\alpha - 6| = |8\alpha - 13|$.
$\frac{|8\alpha - 13|}{\sqrt{16\alpha^2 - 24\alpha + 9 + 4 + 4\alpha^2 - 12\alpha + 9}} = \frac{5}{\sqrt{6}}$.
$\frac{|8\alpha - 13|}{\sqrt{20\alpha^2 - 36\alpha + 22}} = \frac{5}{\sqrt{6}}$.
Squaring both sides: $6(64\alpha^2 - 208\alpha + 169) = 25(20\alpha^2 - 36\alpha + 22)$.
$384\alpha^2 - 1248\alpha + 1014 = 500\alpha^2 - 900\alpha + 550$.
$116\alpha^2 + 348\alpha - 464 = 0$.
Dividing by $116$: $\alpha^2 + 3\alpha - 4 = 0$.
Sum of roots $\alpha_1 + \alpha_2 = -3$.

(D) Let the line $L$ pass through $C(1,1,1)$.
Let the line $L$ intersect the line $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = \lambda$ at point $A(2\lambda+1, 3\lambda-1, 4\lambda+1)$.
Let the line $L$ intersect the line $L_2: \frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1} = \mu$ at point $B(\mu+3, 2\mu+4, \mu)$.
Since $A, B, C$ are collinear,the direction ratios of $AC$ and $BC$ must be proportional.
Direction ratios of $AC$ are $(2\lambda+1-1, 3\lambda-1-1, 4\lambda+1-1) = (2\lambda, 3\lambda-2, 4\lambda)$.
Direction ratios of $BC$ are $(\mu+3-1, 2\mu+4-1, \mu-1) = (\mu+2, 2\mu+3, \mu-1)$.
Since $A, B, C$ are collinear,$\frac{2\lambda}{\mu+2} = \frac{3\lambda-2}{2\mu+3} = \frac{4\lambda}{\mu-1} = k$.
From $\frac{2\lambda}{\mu+2} = \frac{4\lambda}{\mu-1}$,we get $\frac{1}{\mu+2} = \frac{2}{\mu-1} \Rightarrow \mu-1 = 2\mu+4 \Rightarrow \mu = -5$.
Substituting $\mu = -5$ into the coordinates of $B$,we get $B(-5+3, 2(-5)+4, -5) = (-2, -6, -5)$.
The direction ratios of line $L$ (passing through $C(1,1,1)$ and $B(-2, -6, -5)$) are $(1-(-2), 1-(-6), 1-(-5)) = (3, 7, 6)$.
The equation of line $L$ is $\frac{x-1}{3} = \frac{y-1}{7} = \frac{z-1}{6}$.
Checking the options:
For $(7, 15, 13)$: $\frac{7-1}{3} = 2, \frac{15-1}{7} = 2, \frac{13-1}{6} = 2$. Since all ratios are equal,$(7, 15, 13)$ lies on line $L$.

(B) The line $L_1$ is given by $\frac{x-1}{1} = \frac{y-2}{1} = \frac{z-0}{1}$. Let $Q = (\lambda+1, \lambda+2, \lambda)$.
Since $PQ \perp L_1$,the vector $\vec{PQ} = (\lambda+1-5, \lambda+2-1, \lambda-(-3)) = (\lambda-4, \lambda+1, \lambda+3)$ is perpendicular to the direction vector $\vec{v_1} = (1, 1, 1)$.
Thus,$(\lambda-4)(1) + (\lambda+1)(1) + (\lambda+3)(1) = 0 \Rightarrow 3\lambda = 0 \Rightarrow \lambda = 0$.
So,$Q = (1, 2, 0)$ and $\vec{PQ} = (-4, 1, 3)$.
The line $L_2$ is given by $\frac{x-2}{1} = \frac{y-0}{1} = \frac{z-1}{1}$. Let $R = (\mu+2, \mu, \mu+1)$.
Since $PR \perp L_2$,the vector $\vec{PR} = (\mu+2-5, \mu-1, \mu+1-(-3)) = (\mu-3, \mu-1, \mu+4)$ is perpendicular to the direction vector $\vec{v_2} = (1, 1, 1)$.
Thus,$(\mu-3)(1) + (\mu-1)(1) + (\mu+4)(1) = 0 \Rightarrow 3\mu = 0 \Rightarrow \mu = 0$.
So,$R = (2, 0, 1)$ and $\vec{PR} = (-3, -1, 4)$.
The area of $\triangle PQR$ is $A = \frac{1}{2} |\vec{PQ} \times \vec{PR}|$.
$\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 1 & 3 \\ -3 & -1 & 4 \end{vmatrix} = \hat{i}(4 - (-3)) - \hat{j}(-16 - (-9)) + \hat{k}(4 - (-3)) = 7\hat{i} + 7\hat{j} - 7\hat{k}$.
$|\vec{PQ} \times \vec{PR}| = \sqrt{7^2 + 7^2 + (-7)^2} = \sqrt{49 \times 3} = 7\sqrt{3}$.
$A = \frac{1}{2} \times 7\sqrt{3} = \frac{7\sqrt{3}}{2}$.
$4A^2 = 4 \times \frac{49 \times 3}{4} = 147$.

(A) The lines are given by $\vec{r_1} = (1, 2, 3) + t(2, 3, 4)$ and $\vec{r_2} = (\lambda, 4, 5) + s(3, 4, 5)$.
The shortest distance $d$ between two lines $\vec{r} = \vec{a_1} + t\vec{b_1}$ and $\vec{r} = \vec{a_2} + s\vec{b_2}$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$.
Here,$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}$.
The magnitude is $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{6}$.
$\vec{a_2} - \vec{a_1} = (\lambda-1, 2, 2)$.
Shortest distance $d = \frac{|(\lambda-1)(-1) + 2(2) + 2(-1)|}{\sqrt{6}} = \frac{|-\lambda + 1 + 4 - 2|}{\sqrt{6}} = \frac{|3-\lambda|}{\sqrt{6}}$.
Given $d = \frac{1}{\sqrt{6}}$,so $|3-\lambda| = 1$,which gives $\lambda = 4$ or $\lambda = 2$.
Thus,$\lambda_1 = 4$ and $\lambda_2 = 2$.
We need the radius of the circle passing through $(0,0), (4,2)$ and $(2,4)$.
Let the points be $O(0,0), A(4,2), B(2,4)$. The area of $\triangle OAB$ is $\Delta = \frac{1}{2} |0(2-4) + 4(4-0) + 2(0-2)| = \frac{1}{2} |16 - 4| = 6$.
The side lengths are $OA = \sqrt{4^2+2^2} = \sqrt{20}$,$OB = \sqrt{2^2+4^2} = \sqrt{20}$,$AB = \sqrt{(4-2)^2 + (2-4)^2} = \sqrt{8}$.
The radius $R = \frac{abc}{4\Delta} = \frac{\sqrt{20} \cdot \sqrt{20} \cdot \sqrt{8}}{4 \cdot 6} = \frac{20 \cdot 2\sqrt{2}}{24} = \frac{40\sqrt{2}}{24} = \frac{5\sqrt{2}}{3}$.

(C) Let the lines be:
$L_1: x+2=y-1=z=\ell$
$L_2: \frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}=m$
$L_3: \frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}=n$
Point of intersection of $L_1$ and $L_2$:
$x = \ell-2, y = \ell+1, z = \ell$
$x = 5m+3, y = -m, z = m+1$
Equating coordinates: $\ell-2=5m+3, \ell+1=-m, \ell=m+1$. Solving gives $\ell=0, m=-1$. Point $A = (-2, 1, 0)$.
Point of intersection of $L_2$ and $L_3$:
$x = 5m+3, y = -m, z = m+1$
$x = -3n, y = 3n+3, z = n+2$
Equating coordinates: $5m+3=-3n, -m=3n+3, m+1=n+2$. Solving gives $m=0, n=-1$. Point $B = (3, 0, 1)$.
Point of intersection of $L_3$ and $L_1$:
$x = -3n, y = 3n+3, z = n+2$
$x = \ell-2, y = \ell+1, z = \ell$
Equating coordinates: $-3n=\ell-2, 3n+3=\ell+1, n+2=\ell$. Solving gives $\ell=2, n=0$. Point $C = (0, 3, 2)$.
Area of $\triangle ABC = \frac{1}{2} |\vec{AB} \times \vec{AC}|$
$\vec{AB} = (3 - (-2))\hat{i} + (0-1)\hat{j} + (1-0)\hat{k} = 5\hat{i} - \hat{j} + \hat{k}$
$\vec{AC} = (0 - (-2))\hat{i} + (3-1)\hat{j} + (2-0)\hat{k} = 2\hat{i} + 2\hat{j} + 2\hat{k}$
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -1 & 1 \\ 2 & 2 & 2 \end{vmatrix} = \hat{i}(-2-2) - \hat{j}(10-2) + \hat{k}(10+2) = -4\hat{i} - 8\hat{j} + 12\hat{k}$
$Area = \frac{1}{2} \sqrt{(-4)^2 + (-8)^2 + 12^2} = \frac{1}{2} \sqrt{16 + 64 + 144} = \frac{1}{2} \sqrt{224} = \sqrt{56}$
$A^2 = 56$.

(A) First,find the direction vectors of lines $AB$ and $CD$.
The vector $\vec{AB} = (2-1, 3-2, 2-1) = (1, 1, 1)$.
The vector $\vec{CD} = (3-2, 2-1, 4-3) = (1, 1, 1)$.
Since the direction vectors are identical,the lines are parallel.
Now,check if the lines are the same by verifying if point $C(2, 1, 3)$ lies on line $AB$.
The equation of line $AB$ is $\vec{r} = (1, 2, 1) + t(1, 1, 1) = (1+t, 2+t, 1+t)$.
For $C$ to be on $AB$,$(1+t, 2+t, 1+t) = (2, 1, 3)$.
This gives $1+t=2 \implies t=1$,$2+t=1 \implies t=-1$,and $1+t=3 \implies t=2$.
Since $t$ is not consistent,$C$ does not lie on $AB$.
Thus,the lines are parallel but distinct.
Therefore,$\overleftrightarrow{AB} \parallel \overleftrightarrow{CD}$.

(A) Let $AD$ be the median through vertex $A$ to side $BC$.
Since $D$ is the midpoint of $BC$,its coordinates are $D = \left(\frac{\lambda - 1}{2}, \frac{5 + 3}{2}, \frac{\mu + 2}{2}\right) = \left(\frac{\lambda - 1}{2}, 4, \frac{\mu + 2}{2}\right)$.
The vector $\vec{AD}$ is given by $D - A = \left(\frac{\lambda - 1}{2} - 2, 4 - 3, \frac{\mu + 2}{2} - 5\right) = \left(\frac{\lambda - 5}{2}, 1, \frac{\mu - 8}{2}\right)$.
Since the median $AD$ is equally inclined to the coordinate axes,its direction ratios must be equal.
Therefore,$\frac{\lambda - 5}{2} = 1 = \frac{\mu - 8}{2}$.
Solving for $\lambda$: $\frac{\lambda - 5}{2} = 1 \implies \lambda - 5 = 2 \implies \lambda = 7$.
Solving for $\mu$: $\frac{\mu - 8}{2} = 1 \implies \mu - 8 = 2 \implies \mu = 10$.
Thus,$\lambda + \mu = 7 + 10 = 17$.