The perpendicular distance from the origin to the plane $x + 2y - 2z + 5 = 0$ equals $.........$ units.

If the foot of the perpendicular drawn from $(0,0,0)$ to a plane is $(1,2,3)$,then the equation of the plane is:

The distance between the parallel planes $2x - 2y + z + 3 = 0$ and $4x - 4y + 2z + 5 = 0$ is:

The perpendicular distance from the origin to the plane containing the points having position vectors $\hat{i}+2\hat{j}+3\hat{k}$,$2\hat{i}+3\hat{j}-4\hat{k}$,and $3\hat{i}-4\hat{j}+5\hat{k}$ is

If the plane $-4x - 2y + 2z + \alpha = 0$ is at a distance of $2$ units from the plane $2x + y - z + 1 = 0$,then the product of all the possible values of $\alpha$ is

$A$ variable plane passes through a fixed point $(3, 2, 1)$ and meets the $x, y,$ and $z$ axes at $A, B,$ and $C$ respectively. $A$ plane is drawn parallel to the $yz$-plane through $A$,a second plane is drawn parallel to the $zx$-plane through $B$,and a third plane is drawn parallel to the $xy$-plane through $C$. Then,the locus of the point of intersection of these three planes is: