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:

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:

Prove that the lines $x=p y+q, z=r y+s$ and $x=p^{\prime} y+q^{\prime}, z=r^{\prime} y+s^{\prime}$ are perpendicular if $p p^{\prime}+r r^{\prime}+1=0$.

If the lines $\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles,then $p=$

The perpendicular distance from the point $P(-1, 1, 0)$ to the line joining the points $A(0, 2, 4)$ and $B(3, 0, 1)$ is

Let in a $\triangle ABC$,the length of the side $AC$ be $6$,the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ is