Let $ABC$ be a triangle with $A(\alpha, 5, \beta)$, $B(-2, 1, 6)$ and $C(1, 0, -3)$ as its vertices. If the median through $B$ is equally inclined to the coordinate axes, then $\alpha + \beta =$

$A$ line with direction ratios $2, 1, 2$ meets the lines $x = y + 2 = z$ and $x + 2 = 2y = 2z$ at points $P$ and $Q$ respectively. If the length of the perpendicular from the point $(1, 2, 12)$ to the line $PQ$ is $l$,then $l^2$ 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

If $(a, b, c)$ is the image of the point $(1, 2, -3)$ in the line $\frac{x+1}{2} = \frac{y-3}{-2} = \frac{z}{-1}$,then $a+b+c$ is equal to

The lines $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}$ and $\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}$:

The equation of the line passing through $(1, 2, 3)$ and perpendicular to the lines $x-1 = \frac{y+2}{2} = \frac{z+4}{4}$ and $\frac{x-1}{2} = \frac{y-2}{2} = z+3$ is