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

The equation of the plane mid-parallel to the planes $2x - 3y + 6z + 21 = 0$ and $2x - 3y + 6z - 14 = 0$ is given by

The equation of a plane passing through $(-1, 2, 3)$ and whose normal makes equal angles with the coordinate axes is

In space,the equation $by + cz + d = 0$ represents a plane perpendicular to the

If $\overrightarrow{p} = 4\hat{i} - \hat{j} + \hat{k}$ is a point and $\overrightarrow{q} = 9\hat{i} - 2\hat{j} + 6\hat{k}$ is a normal vector,then the perpendicular distance of the origin from the plane passing through $\overrightarrow{p}$ and perpendicular to $\overrightarrow{q}$ is

The coordinates of the foot of the perpendicular drawn from the origin to the plane $2x + y - 2z = 18$ are