Find the coordinates of the foot of the perpendicular drawn from the point $(1, 0, 0)$ to the line $\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z + 10}{8}$.

The direction cosines of a line which is perpendicular to the lines $\frac{x - 7}{2} = \frac{y + 17}{-3} = \frac{z - 6}{1}$ and $\frac{x + 5}{1} = \frac{y + 3}{2} = \frac{z - 6}{-2}$ are...

If the shortest distance between the lines $\frac{x-k}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$,then $k=$

The vertices $B$ and $C$ of a $\Delta ABC$ lie on the line $\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$ such that $BC = 5 \text{ units}$. If the vertex $A$ is $(1, -1, 2)$,then the area of this triangle (in $\text{sq. units}$) is:

The square of the distance of the point $(-2, -8, 6)$ from the line $\frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1}$ along the line $\frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}$ is equal to:

Let the point,on the line passing through the points $P(1, -2, 3)$ and $Q(5, -4, 7)$,farther from the origin and at a distance of $9$ units from the point $P$,be $(\alpha, \beta, \gamma)$. Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to: