$\triangle ABC$ 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

Let a line having direction ratios $1, -4, 2$ intersect the lines $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}$ and $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}$ at the points $A$ and $B$ respectively. Then $( AB )^{2}$ is equal to

If the coordinates of the points $A, B, C,$ and $D$ are $(1, 2, 3), (4, 5, 7), (-4, 3, -6),$ and $(2, 9, 2)$ respectively,then find the angle between the lines $AB$ and $CD$. (in $^\circ$)

$A(1, -2, 1)$ and $B(2, -1, 2)$ are the end points of a line segment. If $D(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from $C(1, 2, 3)$ to $AB$,then $\alpha^2 + \beta^2 + \gamma^2 =$

The length of the perpendicular from the point $(1, 2, 3)$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}$ is

The $xy$-plane divides the line segment joining the points $(2, 4, 5)$ and $(-4, 3, -2)$ in the ratio: