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

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:

The equation of the plane which contains the $y$-axis and passes through the point $(1, 2, 3)$ is:

Let $\pi$ be the plane passing through the point $(3,-3,1)$ and perpendicular to the line joining the points $(3,4,-1)$ and $(2,-1,5)$. If the equation of the plane containing the points $(3,4,-1),(-1,2,5)$ and perpendicular to the plane $\pi$ is $ax+y+cz-d=0$,then $3(a+c)=$

If $a, b, c$ are the intercepts made on $X, Y, Z$-axes respectively by the plane passing through the points $(1, 0, -2), (3, -1, 2)$ and $(0, -3, 4)$, then $3a + 4b + 7c =$

If the foot of the perpendicular drawn from the point $(0, 0, 0)$ to the plane is $(4, -2, -5)$,then the equation of the plane is $......$