If the circumcenter of the triangle formed by the points $(1,2,3), (3,-1,5)$ and $(4,0,-3)$ is $(\alpha, \beta, \gamma)$,then $|\alpha|+|\beta|=$

If the points $(x, -1)$,$(3, y)$,$(-2, 3)$,and $(-3, -2)$ are the vertices of a parallelogram,then:

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

The extremities of a diagonal of a parallelogram are the points $(3, -4)$ and $(-6, 5)$. If the third vertex is $(-2, 1)$,then the fourth vertex 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 the centroid of $\Delta ABC$ is $(0,0,0)$,where $A(1,1,1), B(2,1,2), C(x, y, z)$,then $(x, y, z) = \ldots \ldots$