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 $(2, 2, 1)$.

The equation $xy = 0$ in three-dimensional space represents

If the foot of the perpendicular drawn from the origin to the plane is $(3, 2, 1)$,then the equation of the plane is

In each of the following cases,determine the direction cosines of the normal to the plane and the distance from the origin: $2x + 3y - z = 5$.

$A$ vector $\vec{n}$ is inclined to the $x$-axis at $45^\circ$,to the $y$-axis at $60^\circ$,and at an acute angle to the $z$-axis. If $\vec{n}$ is a normal to a plane passing through the point $(\sqrt{2}, -1, 1)$,then the equation of the plane is:

Equations of planes parallel to the plane $x-2y+2z+4=0$ which are at a distance of $1$ unit from the point $(1, 2, 3)$ are $.....$