If $\alpha$ is the angle between any two diagonals of a cube and $\beta$ is the angle between a diagonal of a cube and a diagonal of its face, which intersects this diagonal of the cube, then $\cos \alpha + \cos^2 \beta =$

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:

The length of the altitude through the point $D$ of a tetrahedron with vertices $A(2,3,1)$,$B(4,1,-2)$,$C(6,3,7)$,and $D(-5,-4,8)$ is: (in $units$)

$A$ line with direction ratios proportional to $2, 1, 2$ meets each of the lines $x = y + a = z$ and $x + a = 2y = 2z$. The coordinates of each of the points of intersection are given by

$A$ point moves such that the sum of its distances from $(4, 0, 0)$ and $(-4, 0, 0)$ is always $10$. The locus of the point 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