The equation of the plane,passing through the midpoint of the line segment joining the points $P(1, 2, 5)$ and $Q(3, 4, 3)$ and perpendicular to it,is

If $A$ and $B$ are the feet of the perpendiculars drawn from the point $Q(a, b, c)$ to the planes $YZ$ and $ZX$ respectively,then the equation of the plane passing through the points $A, B$,and $O$ is (where $O$ is the origin).

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

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

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $2x - 3y + 4z - 6 = 0$.

$A$ plane $x$ passes through the point $(1, 1, 1)$. If $b, c, a$ are the direction ratios of a normal to the plane,where $a, b, c$ $(a < b < c)$ are the factors of $2001$,then the equation of the plane is