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

$A(27, -243, 81)$ is a point in space. $B, C, D$ are images of $A$ with respect to $XY, YZ$, and $ZX$ planes respectively. If the centroid of the triangle $BCD$ is $(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma =$

$AB$ 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 point in the $xy$-plane which is equidistant from the points $A(2,0,3)$,$B(0,3,2)$,and $C(0,0,1)$ has the coordinates

Find the distance of the point $P(-\hat{i} + 2\hat{j} + 6\hat{k})$ from the line passing through the point $A(2, 3, -4)$ and parallel to the vector $\vec{v} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.

The edge of a cube is of length $a$. Then the shortest distance between the diagonal of a cube and an edge skew to it is: