The equation of the plane passing through the point $(1, 2, 1)$ and perpendicular to the planes $x + 2y + 2z - 7 = 0$ and $3x + 3y + 2z - 5 = 0$ is

If $A$ and $B$ are the feet of the perpendiculars drawn from $(1, 2, 3)$ to planes $YZ$ and $ZX$ respectively, then the equation of the plane passing through the points $A$, $B$ and the origin $(0, 0, 0)$ is

$A$ point lying on the plane that passes through the points $\hat{i}-\hat{j}+\hat{k}$, $\hat{i}-2\hat{j}+3\hat{k}$, and $\hat{i}+2\hat{j}-3\hat{k}$ is:

Find the equation of the plane that passes through the three points $(1,1,0), (1,2,1),$ and $(-2,2,-1)$.

If $\overrightarrow{p} = 4\hat{i} - \hat{j} + \hat{k}$ is a point and $\overrightarrow{q} = 9\hat{i} - 2\hat{j} + 6\hat{k}$ is a normal vector,then the perpendicular distance of the origin from the plane passing through $\overrightarrow{p}$ and perpendicular to $\overrightarrow{q}$ is

If the points $(1, 1, k)$ and $(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0$,then $k =$