The length of the perpendicular from the point $A(1, -2, -3)$ on the line $\frac{x-1}{2} = \frac{y+3}{-1} = \frac{z+1}{-2}$ is (in $\text{ units}$)

Find the vector equation of the line which is parallel to the vector $3 \hat{i}-2 \hat{j}+6 \hat{k}$ and which passes through the point $(1, -2, 3)$.

Statement $-1$: The point $A(1, 0, 7)$ is the mirror image of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$.
Statement $-2$: The line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ bisects the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.

Find the vector and the cartesian equations of the line that passes through the points $(3,-2,-5)$ and $(3,-2,6).$

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:

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