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:

$A(2,3,-4), B(-3,3,-2), C(-1,4,2)$ and $D(3,5,1)$ are the vertices of a tetrahedron. If $E, F, G$ are the centroids of its faces containing the point $A$, then the centroid of the triangle $EFG$ is

$A(3,2,-1), B(4,1,1), C(6,2,5)$ 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

Find the reflection of the point $(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$.

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

If the orthocentre and the centroid of a triangle are at $(5,2,-6)$ and $(9,6,-4)$ respectively,then its circumcentre is