The distance of the point $(-2, 4, -5)$ from the line $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$ is

Find the angle between the two lines $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and $\frac{x-1}{10}=\frac{y+3}{2}=\frac{z+4}{-11}$.

$A$ line passes through $A(4, -6, -2)$ and $B(16, -2, 4)$. The point $P(a, b, c)$,where $a, b, c$ are non-negative integers,lies on the line $AB$ at a distance of $21$ units from point $A$. The distance between the points $P(a, b, c)$ and $Q(4, -12, 3)$ is equal to...........

Let $P$ be the point of intersection of the lines $\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$ and $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$. Then,the shortest distance of $P$ from the line $4x=2y=z$ is

Let $P$ be the image of the point $Q(7,-2,5)$ in the line $L: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $R(5, p, q)$ be a point on $L$. Then the square of the area of $\triangle P Q R$ is $\qquad$

$A$ line $L$ passes through points $A(1, 3, 2)$ and $B(2, 2, 1)$. If the mirror image of point $P(1, 1, -1)$ in the line $L$ is $(x, y, z)$,then $x+y+z=$