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

Find the distance of the point $P(3, 8, 2)$ from the line $\frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3}$ measured parallel to the plane $3x + 2y - 2z + 15 = 0$.

What is the angle between two diagonals of a cube?

$A(1,1,1), B(1,-4,3), C(2,-2,0)$ 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

$AB$ and $BC$ are diagonals of adjacent faces of a rectangular box with its center at the origin,and its edges parallel to the coordinate axes. If the angles $\angle BOC, \angle COA$,and $\angle AOB$ are $\alpha, \beta$,and $\gamma$ respectively,then $\cos \alpha + \cos \beta + \cos \gamma$ 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 $units$)