The shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$ and $\frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6}$ is . . . . . . .

Show that the lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other.

An angle between the lines whose direction cosines are given by the equations $l + 3m + 5n = 0$ and $5lm - 2mn + 6nl = 0$ is

The distance of the point $Q(0, 2, -2)$ from the line passing through the point $P(5, -4, 3)$ and perpendicular to the lines $\overrightarrow{r} = (-3 \hat{i} + 2 \hat{k}) + \lambda(2 \hat{i} + 3 \hat{j} + 5 \hat{k}), \lambda \in R$ and $\overrightarrow{r} = (\hat{i} - 2 \hat{j} + \hat{k}) + \mu(-\hat{i} + 3 \hat{j} + 2 \hat{k}), \mu \in R$ is

The line segment joining the points $A(2, 3, 4)$ and $B(-3, 5, -4)$ intersects the $yz$-plane at the point:

Let a line $L$ pass through the point $P(2, 3, 1)$ and be parallel to the line $x + 3y - 2z - 2 = 0 = x - y + 2z$. If the distance of $L$ from the point $(5, 3, 8)$ is $\alpha$,then $3\alpha^2$ is equal to $......$.