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 $\sqrt{3}$)

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 perpendicular, then the value of $p$ is . . . . . . .

The coordinates of the point where the line passing through $A(3, 4, 1)$ and $B(5, 1, 6)$ crosses the $xy$-plane are

The coordinates of the point where the line joining the points $(3, 5, -7)$ and $(-2, 1, 8)$ is intersected by the $yz$-plane are given by:

Let $S$ be the set of all values of $\lambda$,for which the shortest distance between the lines $\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}$ and $\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$ is $13$. Then $8\left|\sum_{\lambda \in S} \lambda\right|$ is equal to

The shortest distance between the lines $\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}$ and $\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}$ is $......$.