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 vertices of a triangle are $(1, 2, 3)$,$(2, 3, 1)$,and $(3, 1, 2)$,and if $H, G, S$,and $I$ respectively denote its orthocenter,centroid,circumcenter,and incenter,then $H+G+S+I$ is equal to:

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

$A(1,1,1), B(1,-4,3), C(2,-2,0)$ and $D(8,1,4)$ are the vertices of a tetrahedron. $G_1, G_2, G_3$ and $G_4$ are the centroids of the faces $ABC, BCD, CDA$ and $DAB$. Then the centroid of the tetrahedron having $G_1, G_2, G_3, G_4$ as its vertices is

Find the reflection of the point $(-1, 3, 4)$ in the plane $x - 2y = 0$.

$A(3,2,-1), B(4,1,1), C(6,2,5)$ 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