If the lines $\frac{x+1}{-10}=\frac{y+k}{-1}=\frac{z-4}{1}$ and $\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}$ intersect each other,then the value of $k$ is

$ABC$ is a triangle 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 $(\lambda^3 + \mu^3 + 5)$ is equal to:

Line $L$ passes through two points $(2, -3, 1)$ and $(3, -4, -5)$. If point $(0, a, b)$ lies on the line $L$,then $a+b =$ . . . . . . .

Find the shortest distance between the lines $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$.

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then the value of $k$ 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