The point collinear with $(1, -2, -3)$ and $(2, 0, 0)$ among the following is

The vector equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines $\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$ and $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$ is . . . . . . .

Find the coordinates of the foot of the perpendicular drawn from the point $P(1, 0, 3)$ to the line joining the points $A(4, 7, 1)$ and $B(3, 5, 3)$.

Find the coordinates of the foot of the perpendicular drawn from the point $A(1, 0, 3)$ to the line joining the points $B(4, 7, 1)$ and $C(3, 5, 3)$.

If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-2}{2}$ is $4$, then the sum of all possible values of $a$ is equal to :

$A$ line passing through the points $P(1, -1, 2)$ and $Q(2, 0, 1)$ meets the $XY$-plane and the $YZ$-plane at points $A$ and $B$ respectively. The distance $AB$ is equal to...