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 $(2, -1, 5)$.

$A$ plane which is perpendicular to two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$ passes through $(1, 2, 1)$. The distance of the plane from the point $(2, 3, 4)$ is

Let $P$ be the plane passing through the point $(1, 2, 3)$ and the line of intersection of the planes $\vec{r} \cdot (\hat{i} + \hat{j} + 4\hat{k}) = 16$ and $\vec{r} \cdot (-\hat{i} + \hat{j} + \hat{k}) = 6$. Then which of the following points does $NOT$ lie on $P$?

The equation of the plane passing through $(-1, 1, 2)$,whose normal makes equal acute angles with the coordinate axes is

The equation of the plane,passing through the point $(1, 1, 1)$ and perpendicular to the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$,is

If the plane $56x + 4y + 9z = 2016$ meets the coordinate axes at points $A, B$,and $C$,then the centroid of the $\triangle ABC$ is