If $O$ is the origin and the coordinates of $P$ are $(1, 2, -3),$ find the equation of the plane passing through $P$ and perpendicular to $OP.$

If the foot of the perpendicular from $(0,0,0)$ to a plane is $(1,2,3)$, then the equation of the plane is

$A$ point on the plane determined by the points $A(1,1,-1)$, $B(2,-1,0)$, and $C(-1,0,2)$ among the following is:

Find the equation of the plane that contains the point $(1, -1, 2)$ and is perpendicular to each of the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$.

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.

$A$ plane $E_{1}$ makes intercepts $1, -3, 4$ on the coordinate axes. The equation of a plane parallel to plane $E_{1}$ and passing through $(2, 6, -8)$ is