Find the angle between the planes $ax + by + d = 0$ $(a^2 + b^2 \neq 0)$ and $z = 0$.

If the planes $3x - 2y + 2z + 17 = 0$ and $4x + 3y - kz = 25$ are mutually perpendicular,then $k = $

If $\alpha$ and $\beta$ are scalars and $\vec{r} = (2+\alpha-3\beta) \hat{i} + (\beta-3) \hat{j} + (2\alpha-5\beta-1) \hat{k}$ is the equation of a plane, then its equation in Cartesian form is:

Let $P$ be the plane passing through the point $(1, -1, -5)$ and perpendicular to the line joining the points $(4, 1, -3)$ and $(2, 4, 3)$. Then the distance of $P$ from the point $(3, -2, 2)$ 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:

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