The shortest distance between the lines $\frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3}$ and $\frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5}$ is:

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{-1} = \frac{z - 3}{1}$ and $\frac{x + 3}{-3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is

$A$ 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...........

Prove that the line through $A(0,-1,-1)$ and $B(4,5,1)$ intersects the line through $C(3,9,4)$ and $D(-4,4,4)$.

The square of the distance of the image of the point $A(6, 1, 5)$ in the line $\frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4}$ from the origin is:

Let $P(\alpha, \beta, \gamma)$ be the image of the point $Q(1, 6, 4)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$. Then $2\alpha + \beta + \gamma$ is equal to ..............