The distance of the point $P(4, 6, -2)$ from the line passing through the point $(-3, 2, 3)$ and parallel to a line with direction ratios $3, 3, -1$ is equal to:

Consider the lines $L_1$ and $L_2$ given by
$L_1: \frac{x-1}{2} = \frac{y-3}{1} = \frac{z-2}{2}$
$L_2: \frac{x-2}{1} = \frac{y-2}{2} = \frac{z-3}{3}$
$A$ line $L_3$ having direction ratios $1, -1, -2$ intersects $L_1$ and $L_2$ at the points $P$ and $Q$ respectively. Then the length of line segment $PQ$ is

The equation of the line passing through the point $(-1, 3, -2)$ and perpendicular to each of the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}$ is

The Cartesian equation of a line is $3x + 1 = 6y - 2 = -z + 1$. Find its vector equation.

If $A(1, 2, 1)$,$B(2, 3, 2)$,$C(2, 1, 3)$,and $D(3, 2, 4)$,then which of the following is true?

The two lines $L_1: \vec{r}=(\hat{i}+5 \hat{j}+5 \hat{k})+t(4 \hat{i}-4 \hat{j}+5 \hat{k})$ and $L_2: \vec{r}=(2 \hat{i}+4 \hat{j}+5 \hat{k})+s(8 \hat{i}-3 \hat{j}+\hat{k})$ are such that