The straight line $\frac{x - 3}{3} = \frac{y - 2}{1} = \frac{z - 1}{0}$ is

The shortest distance between the two lines, where the first line passes through $(0, 0, 0)$ and $(2, 0, 3)$ and the second line passes through $(2, 5, 0)$ and $(0, 4, 0)$, is

If the mirror image of the point $P(3, 4, 9)$ in the line $\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$,then $14(\alpha+\beta+\gamma)$ is :

The square of the distance of the point $\left(\frac{15}{7}, \frac{32}{7}, 7\right)$ from the line $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ in the direction of the vector $\hat{i}+4 \hat{j}+7 \hat{k}$ is :

The perpendicular distance from the point $P(3, -2, 1)$ to the line joining the points $A(1, -3, 5)$ and $B(2, 1, -4)$ is:

Find the foot of the perpendicular from the point $(2, 3, -8)$ to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$. Also,find the perpendicular distance from the given point to the line.