The shortest distance between the lines $r=(-2 \hat{i}+\hat{j}-\hat{k})+r(2 \hat{i}+3 \hat{j}-\hat{k})$ and $r=(\hat{i}-\hat{j}+2 \hat{k})+k(-\hat{i}+2 \hat{j}+4 \hat{k})$ is

Let a line passing through the point $(-1, 2, 3)$ intersect the lines $L_1: \frac{x-1}{3} = \frac{y-2}{2} = \frac{z+1}{-2}$ at $M(\alpha, \beta, \gamma)$ and $L_2: \frac{x+2}{-3} = \frac{y-2}{-2} = \frac{z-1}{4}$ at $N(a, b, c)$. Then the value of $\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}$ equals

If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,then the angle between the lines is

$ABC$ is a triangle in a plane with vertices $A(2, 3, 5)$,$B(-1, 3, 2)$,and $C(\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the coordinate axes,then the value of $\lambda + \mu$ is:

Let $A$ and $B$ be two distinct points on the line $L : \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2\sqrt{17}$ from the foot of the perpendicular drawn from the point $P(1, 2, 3)$ to the line $L$. If $O$ is the origin,then $\overrightarrow{OA} \cdot \overrightarrow{OB}$ is equal to:

The line $L_1$ is parallel to the vector $\vec{a} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and passes through the point $(7, 6, 2)$,and the line $L_2$ is parallel to the vector $\vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and passes through the point $(5, 3, 4)$. The shortest distance between the lines $L_1$ and $L_2$ is: