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

Let $ABCD$ be a tetrahedron such that the edges $AB, AC$ and $AD$ are mutually perpendicular. Let the areas of the triangles $ABC, ACD$ and $ADB$ be $5, 6$ and $7$ square units respectively. Then the area (in square units) of the $\triangle BCD$ is equal to :

Consider the tetrahedron with the vertices $A(3,2,4)$,$B(x_1, y_1, 0)$,$C(x_2, y_2, 0)$,and $D(x_3, y_3, 0)$. If the triangle $BCD$ is formed by the lines $y=x$,$x+y=6$,and $y=1$,then the centroid of the tetrahedron is

If the orthocentre and the centroid of a triangle are $(-3,5,2)$ and $(3,3,4)$ respectively,then its circumcentre is

$A$ tetrahedron of volume $5$ has three of its vertices at the points $A(2,1,-1)$,$B(3,0,1)$,and $C(2,-1,3)$. If the fourth vertex $D$ lies on the $y$-axis,then the sum of the ordinates of all possible points $D$ is-

The equation of the locus of a point whose distance from the $XY$-plane is twice its distance from the $Z$-axis is: