The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{-1} = \frac{z - 3}{1}$ and $\frac{x + 3}{-3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is

If the distances of the point $P(1, 2, a)$ from the line $L: \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $L_{1}: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $L_{2}: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equal, then $a+b+c$ is equal to

Find the vector and the Cartesian equations of the line passing through the point $(5, 2, -4)$ and parallel to the vector $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$.

If the length of the perpendicular drawn from the point $P(a, 4, 2)$,$a > 0$ on the line $\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1}$ is $2\sqrt{6}$ units and $Q(\alpha_{1}, \alpha_{2}, \alpha_{3})$ is the image of the point $P$ in this line,then $a + \sum_{i=1}^{3} \alpha_{i}$ is equal to.

If $P$ is a point lying on the line passing through the point $A(\hat{i}-\hat{j}+3 \hat{k})$ and parallel to the vector $2 \hat{i}+\hat{j}-2 \hat{k}$ such that $|AP|=18$,then a position vector of $P$ is

The lines whose direction cosines satisfy the equations $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$ are perpendicular if: