The perpendicular distance of the point $(1, -1, 2)$ from the plane $x + 2y + z = 4$ is

$A$ plane is parallel to two lines,whose direction ratios are $1, 0, -1$ and $-1, 1, 0$ and it contains the point $(1, 1, 1)$. If it cuts coordinate axes ($X, Y, Z$-axes respectively) at $A, B, C$,then the volume of the tetrahedron $OABC$ is (in cubic units):

If the foot of the perpendicular drawn from $(0,0,0)$ to a plane is $(1,2,3)$,then the equation of the plane is:

If a plane $\pi$ passes through the point $(-1,6,2)$ and is perpendicular to the planes $x+2y+2z-5=0$ and $3x+3y+2z-8=0$, then the perpendicular distance from the point $(1,-1,1)$ to the plane $\pi$ is

The equation of the plane passing through $3 \hat{i}+2 \hat{j}+6 \hat{k}$ and parallel to the vectors $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ is

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