If the three consecutive vertices of a parallelogram are $A(1, 2, 3)$,$B(-1, -2, -1)$,and $C(2, 3, 2)$,then its fourth vertex 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 $\sqrt{6}+\sqrt{13}+\sqrt{11}$

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 $3x + 2y - 2z + 15 = 0$.

$A$ square $ABCD$ of diagonal $2a$ is folded along the diagonal $AC$ so that the planes $DAC$ and $BAC$ are at a right angle. The shortest distance between $DC$ and $AB$ is

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

The angle between two diagonals of a cube is