The distance of the point $P(-2, 4, -5)$ from the line $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$ is

The lines $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$ and $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$ intersect at the point $P$. If the distance of $P$ from the line $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$ is $l$,then $14 l^2$ is equal to:

The sum of the coordinates of one of the points on the line $\frac{x - 2}{1} = \frac{y + 3}{-2} = \frac{z + 5}{2}$ which is at a distance of $3 \text{ units}$ from the point $(2, -3, -5)$ is

The distance of the point $-\hat{i} + 2\hat{j} + 6\hat{k}$ from the straight line that passes through the point $2\hat{i} + 3\hat{j} - 4\hat{k}$ and is parallel to the vector $6\hat{i} + 3\hat{j} - 4\hat{k}$ is

The point of intersection of the lines $l_1: r(t) = (i - 6j + 2k) + t(i + 2j + k)$ and $l_2: R(u) = (4j + k) + u(2i + j + 2k)$ is

Find the coordinates of the foot of the perpendicular from the point $(2, 4, 1)$ to the line $\vec{r} = (-5, -3, 6) + k(1, 4, -9)$,where $k \in R$.