The shortest distance between the lines $\frac{x - 6}{1} = \frac{y - 2}{-2} = \frac{z - 2}{2}$ and $\frac{x + 4}{3} = \frac{y}{-2} = \frac{z + 1}{-2}$ is

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

Find the equation of a line parallel to the $x$-axis and passing through the origin.

Points $P$ and $Q$ are given by $\vec{OP} = \hat{i} - \hat{j} - \hat{k}$ and $\vec{OQ} = -\hat{i} + \hat{j} + \hat{k}$. $A$ line along the vector $\vec{a} = \hat{i} + \hat{j}$ passes through the point $P$ and another line along the vector $\vec{b} = \hat{j} - \hat{k}$ passes through the point $Q$. If a line along the vector $\vec{c} = \hat{i} - \hat{j} + \hat{k}$ intersects both the lines along the vectors $\vec{a}$ and $\vec{b}$ at $L$ and $M$ respectively,then $\vec{PM} =$

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

If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,then the angle between the lines is