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:

$A$ plane $\Pi$ passes through the points $A=(0,0,2)$, $B=(1,0,1)$, and $C=(3,1,1)$. If the plane $\Pi$ makes angles $\alpha$ and $\beta$ with the $XY$ and $XZ$-coordinate planes respectively, then $\sin^2 \alpha + \sin^2 \beta =$

Let $P(1, -2, 5)$ be the foot of the perpendicular drawn from the origin to the plane $\pi_1$ and the same $P$ be the foot of the perpendicular from $(1, 2, -1)$ to the plane $\pi_2$. Then the acute angle between the planes $\pi_1$ and $\pi_2$ is

The equation of a plane parallel to the plane $x - 2y + 2z - 5 = 0$ and at a unit distance from the origin is:

If the length of the perpendicular drawn from the origin to a plane is $7$ units and its direction ratios are $-3, 2, 6$,then the equation of the plane is:

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