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

$A$ point on the plane determined by the points $A(1,1,-1)$, $B(2,-1,0)$, and $C(-1,0,2)$ among the following is:

Let $P$ be a plane passing through the points $(2,1,0)$,$(4,1,1)$,and $(5,0,1)$,and $R$ be the point $(2,1,6)$. Then the image of $R$ in the plane $P$ is:

The plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ cuts the coordinate axes at the points $A, B, C$ respectively. Then the area of triangle $ABC$ is

$A$ plane passes through $(1, -2, 1)$ and is perpendicular to the planes $2x - 2y + z = 0$ and $x - y + 2z = 4$. The distance of the point $(1, 2, 2)$ from this plane is . . . . . . units.

The distance of the point $(1,3,-7)$ from the plane passing through the point $(1,-1,-1)$ having normal perpendicular to both the lines $\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$ is