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

$ABCD$ is a parallelogram,$P$ is the mid-point of $AB$. If $R$ is the point of intersection of $AC$ and $DP$,then $R$ divides $AC$ internally in the ratio

Given $\triangle ABC$ such that $A = 2\hat{i} - \hat{j} + \hat{k}$,$B = \hat{i} - 3\hat{j} - 5\hat{k}$,and $C = 3\hat{i} - 4\hat{j} - 4\hat{k}$,then $\triangle ABC$ is:

Three vertices of a parallelogram $ABCD$ are $A(3, -1, 2)$,$B(1, 2, -4)$,and $C(-1, 1, 2)$. Find the coordinates of the fourth vertex $D$.

$\triangle ABC$ has vertices at $A \equiv (2, 3, 5)$,$B \equiv (-1, 3, 2)$ and $C \equiv (\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the axes,then the values of $\lambda$ and $\mu$ respectively are

If $A(3, 5), B(-5, -4), C(7, 10)$ are the vertices of a parallelogram,taken in order,then the coordinates of the fourth vertex are: