The incentre of the triangle whose vertices are $P(0,3,0)$,$Q(0,0,4)$,and $R(0,3,4)$ is

What is the angle between two diagonals of a cube?

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

$A$ 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 find the value of $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma + \sin^2 \delta$.

Let $ABCD$ be a tetrahedron in which the coordinates of each of its vertices are in arithmetic progression with the same common difference. If the centroid $G$ of the tetrahedron is $(2, 3, k)$,then the distance of $G$ from the origin is