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 $D$.

If two vertices of a triangle are $A(3,1,4)$ and $B(-4,5,-3)$ and the centroid of the triangle is $G(-1,2,1)$,then the third vertex $C$ of the triangle is

Let $A(x, y, z)$ be a point in the $xy$-plane,which is equidistant from three points $P(0, 3, 2)$,$Q(2, 0, 3)$,and $R(0, 0, 1)$. Let $B = (1, 4, -1)$ and $C = (2, 0, -2)$. Then among the statements $(S1) :$ $\triangle ABC$ is an isosceles right-angled triangle and $(S2) :$ the area of $\triangle ABC$ is $\frac{9 \sqrt{2}}{2}$.

If $A(-4, 5, p)$,$B(3, 1, 4)$,and $C(-2, 0, q)$ are the vertices of a triangle $ABC$ and $G(r, q, 1)$ is its centroid,then the value of $2p + q - r$ is equal to

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

If the origin is the centroid of the triangle for which $(-2, 3, 4)$ and $(3, -1, 5)$ are two vertices,then the third vertex is