The circumcentre of the triangle formed by the points $(1, 2, 3), (3, -1, 5), (4, 0, -3)$ is

If the centroid of the triangle whose vertices are $(a, 1, 3)$,$(-2, b, -5)$,and $(4, 7, c)$ is the origin,then $a^2 + b^2 + c^2 =$

If $D(2, 1, 0)$,$E(2, 0, 0)$,and $F(0, 1, 0)$ are mid-points of the sides $BC$,$CA$,and $AB$ of $\triangle ABC$,respectively,then the centroid of $\triangle ABC$ is

The midpoints of the sides of a triangle are $(1, 5, -1)$,$(0, 4, -2)$,and $(2, 3, 4)$. Find the centroid of the triangle.

If $G(3, -5, r)$ is the centroid of $\triangle ABC$,where $A \equiv (7, -8, 1)$,$B \equiv (p, q, 5)$,and $C \equiv (q+1, 5p, 0)$ are vertices of the triangle $ABC$,then the values of $p, q, r$ are respectively:

If the points $P = \hat{i} + 2 \hat{j}$,$Q = 4 \hat{i} + 6 \hat{j}$,$R = 5 \hat{i} + 7 \hat{j}$,and $S = a \hat{i} + b \hat{j}$ are the consecutive vertices of a parallelogram $PQRS$,then: