The centroid of a tetrahedron with vertices $P(5, -7, 0)$,$Q(a, 5, 3)$,$R(4, -6, b)$,and $S(6, c, 2)$ is $(4, -3, 2)$. Then the value of $2a + 3b + c$ is equal to:

If $\left(\frac{9}{4}, \frac{5}{4}, \frac{15}{4}\right)$ is the centroid of a tetrahedron whose vertices are $(a, 2, 1), (1, b, 4), (4, 0, c)$ and $(1, 1, 7)$, then

Let $A(1, 2, 3)$, $B(-1, 4, 6)$, $C(0, -6, 4)$, and $D(1, 1, 1)$ be the vertices of a tetrahedron. Let $G$ be its centroid and $G_1$ be the centroid of its face $BCD$. Then $\frac{AG_1}{AG} =$

$A$ point moves such that the sum of its distances from $(4, 0, 0)$ and $(-4, 0, 0)$ is always $10$. The locus of the point is:

Find the length of the perpendicular drawn from the point $P(2\hat{i} - \hat{j} + 5\hat{k})$ to the line $\vec{r} = (11\hat{i} - 2\hat{j} - 8\hat{k}) + \lambda(10\hat{i} - 4\hat{j} - 11\hat{k})$.

$\Pi_1, \Pi_2, \Pi_3$ 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