In the following case,find the distance of the given point from the corresponding given plane.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

$A$ 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:

If the plane $2x - y + 2z + 3 = 0$ has distances of $\frac{1}{3}$ and $\frac{2}{3}$ units from the planes $4x - 2y + 4z + \lambda = 0$ and $2x - y + 2z + \mu = 0$ respectively,then the maximum value of $\lambda + \mu$ is equal to:

The equation of the plane passing through the points $(1, 2, 3)$,$(-1, 4, 2)$ and $(3, 1, 1)$ is

In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angles between them.
$2x + y + 3z - 2 = 0$ and $x - 2y + 5 = 0$

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