If $(a, b, c)$ is the image of the point $(1, 2, -3)$ in the line $\frac{x+1}{2} = \frac{y-3}{-2} = \frac{z}{-1}$,then $a+b+c$ is equal to

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.

The shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$ is (in $\sqrt{3}$)

Show that the points $A(1, 2, 7)$,$B(2, 6, 3)$,and $C(3, 10, -1)$ are collinear.

If the line joining the points $A(2, 3, -1)$ and $B(3, 5, -3)$ is perpendicular to the line joining the points $C(1, 2, 3)$ and $D(3, y, 7)$,then $y=$

The perpendicular distance of the point $(2, 4, -1)$ from the line $\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9}$ is