Find the values of $p$ so that 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.

If a point $R(4, y, z)$ lies on the line segment joining the points $P(2, -3, 4)$ and $Q(8, 0, 10)$,then the distance of $R$ from the origin is

If the coordinates of the points $A, B, C, D$ are $(1, 2, 3), (4, 5, 7), (-4, 3, -6)$ and $(2, 9, 2)$ respectively,then the angle between the lines $AB$ and $CD$ is

Let $ABC$ be a triangle with $A(\alpha, 5, \beta)$, $B(-2, 1, 6)$ and $C(1, 0, -3)$ as its vertices. If the median through $B$ is equally inclined to the coordinate axes, then $\alpha + \beta =$

Statement-$1$: The shortest distance between the skew lines $\frac{x+3}{-4} = \frac{y-6}{3} = \frac{z}{2}$ and $\frac{x+3}{-4} = \frac{y}{1} = \frac{z-7}{1}$ is $9$.
Statement-$2$: Two lines are skew lines if there exists no plane passing through them.

If a point $R(4, y, z)$ lies on the line joining the points $P(2, -3, 4)$ and $Q(8, 0, 10)$,then the distance of $R$ from the origin is