The distance of the point $A(7, -2, 11)$ from the line $\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$ measured along the line $\frac{x-7}{2} = \frac{y+2}{-3} = \frac{z-11}{6}$ is:

The perpendicular distance from the point $P(-1, 1, 0)$ to the line joining the points $A(0, 2, 4)$ and $B(3, 0, 1)$ is

Show that the points $A(1, 2, 7)$,$B(2, 6, 3)$,and $C(3, 10, -1)$ are collinear.

If the two lines $l_{1}: \frac{x-2}{3}=\frac{y+1}{-2}, z=2$ and $l_{2}: \frac{x-1}{1}=\frac{2y+3}{\alpha}=\frac{z+5}{2}$ are perpendicular,then the angle between the lines $l_{2}$ and $l_{3}: \frac{1-x}{3}=\frac{2y-1}{-4}=\frac{z}{4}$ is

Let $A(2,3,5), B(-1,3,2), C(\lambda, 5, \mu)$ be the vertices of $\triangle ABC$. If the median through the vertex $A$ is equally inclined to the coordinate axes,then

The line passing through the points $(5, 1, a)$ and $(3, b, 1)$ crosses the $yz$-plane at $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$. Then the value of $2a + 3b$ is: