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 foot of the perpendicular from the point $(1, 2, 3)$ on the line $\vec{r} = (6 \hat{i} + 7 \hat{j} + 7 \hat{k}) + \lambda(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ has the coordinates:

If the mirror image of the point $P(3, 4, 9)$ in the line $\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$,then $14(\alpha+\beta+\gamma)$ is :

The equation of a line in Cartesian form passing through $(0, 0, 0)$ and $(4, 3, c)$ and parallel to $a \times b$ where $a = 2i + j + 2k$ and $b = 3i - 4j$ is given by the direction vector $a \times b$. Find the value of $c$ and the equation of the line.

Show that the line through the points $(1, -1, 2)$ and $(3, 4, -2)$ is perpendicular to the line through the points $(0, 3, 2)$ and $(3, 5, 6)$.

Find the length of the perpendicular drawn from the point $P(3, -1, 11)$ to the line $\frac{x}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$.