If $A(0,1,2)$,$B(2,-1,3)$,and $C(1,-3,1)$ are the vertices of a triangle,then the distance between its circumcentre and orthocentre 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 $\sqrt{6}+\sqrt{13}+\sqrt{11}$

$\Pi_1, \Pi_2, \Pi_3$ are three planes which are respectively parallel to the $YZ, ZX$ and $XY$ planes at distances $a, b$ and $c$ forming a rectangular parallelopiped. $d_1$ is a diagonal of the face of $XY$-plane not passing through the origin and $d_2$ is a diagonal of the plane $\Pi_2$ coterminous with $d_1$. If none of the coordinates of the vertices of the parallelopiped are negative, then the angle between $d_1$ and $d_2$ is

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

Find the reflection of the point $P(2, -1, 3)$ in the plane $3x - 2y - z = 9$.

$A(3,2,-1), B(4,1,1), C(6,2,5)$ and $D(3,3,3)$ are four points. $G_1, G_2, G_3$ and $G_4$ are the centroids of the triangles $\triangle BCD, \triangle CDA, \triangle DAB$ and $\triangle ABC$ respectively. The point of concurrence of the lines $AG_1, BG_2, CG_3$ and $DG_4$ is