Let $P$ be the point $(10, -2, -1)$ and $Q$ be the foot of the perpendicular drawn from the point $R(1, 7, 6)$ on the line passing through the points $(2, -5, 11)$ and $(-6, 7, -5)$. Then the length of the line segment $PQ$ is equal to ..........

The lines $\frac{x - 1}{2} = \frac{y + 1}{2} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - 6}{2} = \frac{z}{1}$ intersect each other at point

The lines $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$ and $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$ intersect at the point $P$. If the distance of $P$ from the line $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$ is $l$,then $14 l^2$ is equal to:

The equation of the line passing through the point $(3,1,2)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$ is

The point of intersection of the lines $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ and $\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}$ is

If the length of the perpendicular drawn from the point $P(a, 4, 2)$,$a > 0$ on the line $\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1}$ is $2\sqrt{6}$ units and $Q(\alpha_{1}, \alpha_{2}, \alpha_{3})$ is the image of the point $P$ in this line,then $a + \sum_{i=1}^{3} \alpha_{i}$ is equal to.