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 $\alpha + \beta =$

The length of the perpendicular drawn from the point $(3, -1, 11)$ to the line $\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ is

If the line joining points $(2, 1, 4)$ and $(a - 1, 4, -1)$ is parallel to the line joining points $(0, 2, b - 1)$ and $(5, 3, -2)$, then the values of $a$ and $b$ are respectively:

Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$. Then $(\alpha-\beta)^2$ is equal to ....................

The shortest distance between the lines $\vec{r} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k})$ and $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(\hat{i} + 4\hat{j} - 5\hat{k})$ is:

The angle between two lines $\frac{x+3}{2}=\frac{-y}{3}=\frac{z+5}{-6}$ and $\frac{x-1}{10}=\frac{y+1}{-2}=\frac{z-3}{11}$ is . . . . . . .