The length of the internal bisector of angle $A$ in $\triangle ABC$ with vertices $A(4,7,8)$,$B(2,3,4)$ and $C(2,5,7)$ is

$A(3,2,-1), B(4,1,1), C(6,2,5)$ and $D(3,3,3)$ are four points. $G_1, G_2, G_3$ and $G_4$ are the centroids of the triangles $\triangle BCD, \triangle CDA, \triangle DAB$ and $\triangle ABC$ respectively. The point of concurrence of the lines $AG_1, BG_2, CG_3$ and $DG_4$ is

The image point of $(5, 4, 6)$ in the plane $x + y + 2z - 15 = 0$ is

Let $A=(1,2,0)$, $B=(2,0,-1)$, $C=(0,-2,3)$ and $D=(-1,2,-3)$ be four points in space. Let $G_1$ be the centroid of triangle $ABC$ and $G_2$ be the centroid of tetrahedron $ABCD$. If $P$ divides $G_1G_2$ in the ratio $4:3$ internally, then $P=$

The incentre of the triangle $ABC$,whose vertices are $A(0,2,1)$,$B(-2,0,0)$ and $C(-2,0,2)$,is

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