In three-dimensional space,if $f(x, y, z) = xy + xz$,then the locus of all points which satisfy the equation $f(x, y, z) = 0$ is -

Find the equation of the plane passing through the points $A(2, 2, -1)$,$B(3, 4, 2)$,and $C(7, 0, 6)$.

If the product of the distances of the point $(1, 2, 3)$ from the origin and the plane $2x - 3y + z + k = 0$ is $7$, then the value of $k$ is

If from a point $P(a, b, c)$ perpendiculars $PA$ and $PB$ are drawn to the $yz$-plane and $zx$-plane respectively,then the equation of the plane $OAB$ (where $O$ is the origin) is:

$A$ plane $\pi_1$ passing through the point $3 \hat{i}-7 \hat{j}+5 \hat{k}$ is perpendicular to the vector $\hat{i}+2 \hat{j}-2 \hat{k}$ and another plane $\pi_2$ passing through the point $2 \hat{i}+7 \hat{j}-8 \hat{k}$ is perpendicular to the vector $3 \hat{i}+2 \hat{j}+6 \hat{k}$. If $p_1$ and $p_2$ are the perpendicular distances from the origin to the planes $\pi_1$ and $\pi_2$ respectively, then $p_1-p_2=$

The equations of planes parallel to the plane $x+2y+2z+8=0$,which are at a distance of $2$ units from the point $(1,1,2)$ are