$A$ plane passes through $(1, -2, 1)$ and is perpendicular to the planes $2x - 2y + z = 0$ and $x - y + 2z = 4$. The distance of the point $(1, 2, 2)$ from this plane is . . . . . . units.

Consider three planes:
$P_1: x-y+z=1$
$P_2: x+y-z=-1$
$P_3: x-3y+3z=2$
Let $L_1, L_2, L_3$ be the lines of intersection of the planes $P_2$ and $P_3$,$P_3$ and $P_1$,and $P_1$ and $P_2$,respectively.
$STATEMENT-1$: At least two of the lines $L_1, L_2$ and $L_3$ are non-parallel.
$STATEMENT-2$: The three planes do not have a common point.

If $A$ and $B$ are the feet of the perpendiculars drawn from the point $Q(a, b, c)$ to the planes $YZ$ and $ZX$ respectively,then the equation of the plane passing through the points $A, B$,and $O$ is (where $O$ is the origin).

If for a plane,the intercepts on the coordinate axes are $8, 4, 4$,then the length of the perpendicular from the origin onto the plane is:

The equation of the plane passing through $(1, 1, 1)$ and $(1, -1, -1)$ and perpendicular to $2x - y + z + 5 = 0$ is:

The value of $k$ for which the planes $3x - 6y - 2z = 7$ and $2x + y - kz = 5$ are perpendicular to each other,is