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

$A$ plane passes through $(2,3,-1)$ and is perpendicular to the line having direction ratios $3,-4,7$. The perpendicular distance from the origin to this plane is

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 vector and cartesian equations of the plane which passes through the point $(5, 2, -4)$ and is perpendicular to the line with direction ratios $2, 3, -1$.

If planes $\vec{r} \cdot (p \hat{i} - \hat{j} + 2 \hat{k}) + 3 = 0$ and $\vec{r} \cdot (2 \hat{i} - p \hat{j} - \hat{k}) - 5 = 0$ include an angle $\frac{\pi}{3}$,then the value of $p$ is

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 $(2, 2, 1)$.