Find the length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}$.

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

Find the angle between the lines $\vec{r}=3 \hat{i}-2 \hat{j}+6 \hat{k}+\lambda(2 \hat{i}+\hat{j}+2 \hat{k})$ and $\vec{r}=(2 \hat{j}-5 \hat{k})+\mu(6 \hat{i}+3 \hat{j}+2 \hat{k})$.

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ 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:

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$,and the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}$ is $\alpha$,then $28 \alpha^2$ is equal to $........$.