If the foot of the perpendicular drawn from the point $A(1, 0, 3)$ on a line passing through $B(\alpha, 7, 1)$ is $P\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$,then $\alpha$ is equal to:

Let $L$ be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let $S$ be the set of all points $(a, b, c)$ on $L$, whose distance from the point $P(-1, -1, -9)$ is $7$. Then $\sum_{(a,b,c)\in S} (a+b+c)$ is equal to:

The equation of the line passing through the point of intersection of $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ and also through the point $(2,1,-2)$ is

Find the foot of the perpendicular from the point $(2, 3, -8)$ to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$. Also,find the perpendicular distance from the given point to the line.

Find the equation of the line passing through the point $P(2, -3, 1)$ and perpendicular to the line $L: \frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{-1}$.

The square of the distance of the point $P(5,6,7)$ from the line $\frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4}$ is equal to: