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

The point of intersection of the lines $l_1: r(t) = (i - 6j + 2k) + t(i + 2j + k)$ and $l_2: R(u) = (4j + k) + u(2i + j + 2k)$ is

Let the foot of the perpendicular from the point $(\lambda, 2, 3)$ on the line $\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$ be the point $(1, \mu, 2)$. Then the distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$ and $\frac{x-\lambda}{2} = \frac{y-\mu}{3} = \frac{z+5}{6}$ is equal to:

The distance of the point $P(-3, 2, 3)$ from the line passing through $A(4, 6, -2)$ with direction ratios $\langle -1, 2, 3 \rangle$ is . . . . . . units.

The graph of the equation $y^2 + z^2 = 0$ in three-dimensional space is

The shortest distance between the lines $1+x=2y=-12z$ and $x=y+2=6z-6$ is