Consider the following three planes:
$P : x + y - 2z + 7 = 0$
$Q : x + y + 2z + 2 = 0$
$R : 3x + 3y - 6z - 11 = 0$

The perpendicular distance of the origin from the plane $x-3y+4z-6=0$ is

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

Find the equation of the plane passing through the intersection of the planes $3x - y + 2z - 4 = 0$ and $x + y + z - 2 = 0$ and the point $(2, 2, 1)$.

$A$ variable plane passes through the fixed point $(3, 2, 1)$ and meets $X, Y,$ and $Z$ axes at points $A, B,$ 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:

The direction cosines of a normal to the plane passing through $(4,2,3)$,$(-1,4,2)$ and $(3,2,1)$ are .....