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

Find the plane perpendicular to both given planes and passing through $(1, -2, 1)$.
The required plane's normal vector $\mathrm{n}$ is perpendicular to normals $\mathrm{n_1} = (2, -2, 1)$ and $\mathrm{n_2} = (1, -1, 2)$, so $\mathrm{n} = \mathrm{n_1} \times \mathrm{n_2} = ( -3, -3, 0 )$, which simplifies to $(1, 1, 0)$.

Using point-normal form with $(1, -2, 1)$: $1(x - 1) + 1(y + 2) + 0(z - 1) = 0$ simplifies to $x + y + 1 = 0$.

Compute the distance from point $(1, 2, 2)$ to the plane.
The distance formula gives $d = \frac{|1 \cdot 1 + 1 \cdot 2 + 0 \cdot 2 + 1|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.

$2\sqrt{2}$ corresponds to option C.