If the plane passing through the points $(1, 2, 3)$, $(2, 3, 1)$, and $(3, 1, 2)$ is $a x + b y + c z = 1$, then $a + 2 b + 3 c = $

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

The equation of the plane in normal form which passes through the points $(-2,1,3), (1,1,1)$ and $(2,3,4)$ is

Find the Cartesian equation of the plane passing through the point $(3, -3, 1)$ and normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$.

Assertion: The points $(2, 1, 5)$ and $(3, 4, 3)$ lie on opposite sides of the plane $2x + 2y - 2z - 1 = 0$.
Reason: The algebraic perpendicular distances from the given points to the plane have opposite signs.

The equation of the plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by: