The shortest distance between the lines $r = (3i - 2j - 2k) + t(i)$ and $r = (i - j + 2k) + s(j)$ ($t$ and $s$ being parameters) is

The point collinear with $(1, -2, -3)$ and $(2, 0, 0)$ among the following is

$\triangle ABC$ is formed by $A(1, 8, 4)$,$B(0, -11, 4)$,and $C(2, -3, 1)$. If $D$ is the foot of the perpendicular drawn from $A$ to $BC$,then the coordinates of $D$ are

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:

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