Let the line $L$ intersect the lines $x-2=-y=z-1$ and $2(x+1)=2(y-1)=z+1$,and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$. Then which of the following points lies on $L$?

If $(a, b, c)$ are the direction ratios of a line joining the points $(4, 3, -5)$ and $(-2, 1, -8)$,then the point $P(a, 3b, 2c)$ lies on the plane:

When are the two lines $x = ay + b, z = cy + d$ and $x = a'y + b', z = c'y + d'$ perpendicular to each other?

Let $A$ be the point of intersection of the lines $L_1: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}$ and $L_2: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$. Let $B$ and $C$ be points on the lines $L_1$ and $L_2$ respectively such that $AB = AC = \sqrt{15}$. Then the square of the area of the triangle $ABC$ is:

Let $l_1$ be the line passing through the point $A = 3\hat{i} + 4\hat{j} - 2\hat{k}$ and parallel to the vector $\vec{b_1} = -\hat{i} + 2\hat{j} + \hat{k}$. Let $l_2$ be another line passing through the point $B = \hat{i} - 7\hat{j} - 2\hat{k}$ and parallel to the vector $\vec{b_2} = \hat{i} + 3\hat{j} + 2\hat{k}$. Then the shortest distance between the lines $l_1$ and $l_2$ is:

$L_1$ is a line passing through the points with position vectors $\hat{i}-2 \hat{j}-\hat{k}$ and $4 \hat{i}-3 \hat{k}$. $L_2$ is a line passing through the points with position vectors $\hat{i}+2 \hat{j}-\hat{k}$ and $2 \hat{i}-4 \hat{j}-5 \hat{k}$. Then the distance between $L_1$ and $L_2$ is