What is the angle between two diagonals of a cube?

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

The equation of the locus of a point whose distance from the $XY$-plane is twice its distance from the $Z$-axis is:

$A(1,1,1), B(1,-4,3), C(2,-2,0)$ and $D(8,1,4)$ are the vertices of a tetrahedron. $G_1, G_2, G_3$ and $G_4$ are the centroids of the faces $ABC, BCD, CDA$ and $DAB$. Then the centroid of the tetrahedron having $G_1, G_2, G_3, G_4$ as its vertices is

If the vertices of $\Delta ABC$ are $(a, 0, 0)$,$(0, b, 0)$,and $(0, 0, c)$ respectively,then $\angle B = \dots$

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 $units$)