The distance of the point $(-2, 4, -5)$ from the line $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$ is

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 $......$.

If lines $\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ are mutually perpendicular to each other,then $p = $ . . . . . . .

The point of intersection of the lines represented by $\overline{r}=(\overline{i}-6 \overline{j}+2 \overline{k})+t(\overline{i}+2 \overline{j}+\overline{k})$ and $\overline{r}=(4 \overline{j}+\overline{k})+s(2 \overline{i}+\overline{j}+2 \overline{k})$ is

The shortest distance between lines $L_1$ and $L_2$,where $L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$ and $L_2$ is the line passing through the points $A(-4,4,3)$ and $B(-1,6,3)$ and perpendicular to the line $\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$,is

If two lines $L_1$ and $L_2$ in space are defined by $L_1 = \{ x = \sqrt{\lambda} y + (\sqrt{\lambda} - 1), z = (\sqrt{\lambda} - 1)y + \sqrt{\lambda} \}$ and $L_2 = \{ x = \sqrt{\mu} y + (1 - \sqrt{\mu}), z = (1 - \sqrt{\mu})y + \sqrt{\mu} \}$,then $L_1$ is perpendicular to $L_2$ for all non-negative reals $\lambda$ and $\mu$ such that: