The coordinates of the foot of the perpendicular drawn from a point $\mathrm{P}(-1,1,2)$ to the plane $2 x-3 y+z-11=0$

The foot of the perpendicular from point $P(-1,1,2)$ to the plane $2x - 3y + z - 11 = 0$ is determined by its intersection with the line through $P$ parallel to the plane's normal vector $\mathrm{n} = (2,-3,1)$.

Using the parametric form $\frac{x+1}{2} = \frac{y-1}{-3} = \frac{z-2}{1} = k$, the coordinates of a general point on this line are $x = 2k - 1$, $y = -3k + 1$, $z = k + 2$. Substituting into the plane equation gives $2(2k - 1) - 3(-3k + 1) + (k + 2) - 11 = 0$, which simplifies to $14k - 14 = 0$, yielding $k = 1$.

The coordinates at $k=1$ are $(1,-2,3)$, matching option C.

Final answer: $\boxed{\text{C}}$