$A$ 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-

$A$ 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 $\sqrt{6}+\sqrt{13}+\sqrt{11}$

If a line makes angles $\alpha, \beta, \gamma, \delta$ with the four diagonals of a cube,then the value of $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = $

If the vertices of a triangle are $(1, 2, 3)$,$(2, 3, 1)$,and $(3, 1, 2)$,and if $H, G, S$,and $I$ respectively denote its orthocenter,centroid,circumcenter,and incenter,then $H+G+S+I$ is equal to:

Consider the tetrahedron with the vertices $A(3,2,4)$,$B(x_1, y_1, 0)$,$C(x_2, y_2, 0)$,and $D(x_3, y_3, 0)$. If the triangle $BCD$ is formed by the lines $y=x$,$x+y=6$,and $y=1$,then the centroid of the tetrahedron is

If $P \equiv (0, 1, 0)$ and $Q \equiv (0, 0, 1)$,then what is the projection of $PQ$ on the plane $x + y + z = 3$?