$A$ tetrahedron of volume $5$ has three of its vertices at the points $A(2,1,-1)$,$B(3,0,1)$,and $C(2,-1,3)$. If the fourth vertex $D$ lies on the $y$-axis,then the sum of the ordinates of all possible points $D$ is-

The length of the altitude through the point $D$ of a tetrahedron with vertices $A(2,3,1)$,$B(4,1,-2)$,$C(6,3,7)$,and $D(-5,-4,8)$ is: (in $units$)

$A$ straight line drawn from the point $P(1,3,2)$,parallel to the line $\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}$,intersects the plane $L_1: x-y+3z=6$ at the point $Q$. Another straight line which passes through $Q$ and is perpendicular to the plane $L_1$ intersects the plane $L_2: 2x-y+z=-4$ at the point $R$. Then which of the following statements is(are) $TRUE$?
$(A)$ The length of the line segment $PQ$ is $\sqrt{6}$
$(B)$ The coordinates of $R$ are $(1,6,0)$
$(C)$ The centroid of the triangle $PQR$ is $\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right)$
$(D)$ The perimeter of the triangle $PQR$ is $\sqrt{6}+\sqrt{13}+\sqrt{11}$

$A$ line with direction ratios $2, 1, 2$ intersects the lines $x = y + a = z$ and $x + a = 2y = 2z$. Find the coordinates of the points of intersection.

If $\left(\frac{9}{4}, \frac{5}{4}, \frac{15}{4}\right)$ is the centroid of a tetrahedron whose vertices are $(a, 2, 1), (1, b, 4), (4, 0, c)$ and $(1, 1, 7)$, then

Find the distance of the point $P(3, 8, 2)$ from the line $\frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3}$ measured parallel to the plane $3x + 2y - 2z + 15 = 0$.