If $A(-1, 2, 3)$,$B(1, 1, 1)$ and $C(2, -1, 3)$ are points on a plane,then a unit normal vector to the plane $ABC$ is:

In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angle between them: $7x + 5y + 6z + 30 = 0$ and $3x - y - 10z + 4 = 0$.

The equations $|x| = p, |y| = p, |z| = p$ in $xyz$ space represent:

$A$ 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 mirror image of the point $P(-1, 2, -4)$ in the plane $x - y - 2z + 1 = 0$ is

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