A plane passes through $(2,1,2)$ and $(1,2,1)$ and parallel to the line $2 x=3 y$ and $\mathrm{z}=1$, then the plane also passes through the point.

The plane contains points $P_1(2,1,2)$ and $P_2(1,2,1)$, and is parallel to the line defined by $2x=3y$ and $z=1$.

A vector in the plane is given by $\vec{v}_1 = P_2 - P_1 = (-1, 1, -1)$.

The direction vector of the line is $\vec{d} = (3, 2, 0)$, obtained by parameterizing $x = 3t$, $y = 2t$, $z = 1$ from the line equations.

The normal vector to the plane is perpendicular to both $\vec{v}_1$ and $\vec{d}$, computed as their cross product:

$\vec{n} = \vec{v}_1 \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 1 & -1 \\ 3 & 2 & 0 \end{vmatrix} = (2, -3, -5)$.

Using point $P_1(2,1,2)$ and normal vector $(2, -3, -5)$, the plane equation becomes $2(x-2) - 3(y-1) - 5(z-2) = 0$, which simplifies to $2x - 3y - 5z + 9 = 0$.

Evaluating the given options: only $(-2,0,1)$ satisfies $2(-2) - 3(0) - 5(1) + 9 = 0$.

Answer: $\boxed{C}$