The equation of the plane passing through $(-2, 2, 2)$ and $(2, -2, -2)$ and perpendicular to the plane $9x - 13y - 3z = 0$ is

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

The point which lies on the plane passing through the points $A(\hat{i}-2\hat{j}-3\hat{k})$,$B(3\hat{i}-\hat{j}+4\hat{k})$,and $C(-3\hat{i}+2\hat{j}-5\hat{k})$ is:

The equation of the plane passing through a point $A(2, -1, 3)$ and parallel to the vectors $\vec{a} = (3, 0, -1)$ and $\vec{b} = (-3, 2, 2)$ is:

$A$ plane $\pi$ passing through the point $(1,1,1)$ is perpendicular to the line joining the points $(6,3,2)$ and $(1,-4,-9)$. If $ax+by+cz-23=0$ is the equation of the plane $\pi$,then $a+b-c=$

$A$ variable plane passes through the fixed point $(3, 2, 1)$ and meets $X, Y,$ and $Z$ axes at points $A, B,$ and $C$ respectively. $A$ plane is drawn parallel to the $YZ$-plane through $A$,a second plane is drawn parallel to the $ZX$-plane through $B$,and a third plane is drawn parallel to the $XY$-plane through $C$. Then the locus of the point of intersection of these three planes is: