If $\mathrm{f}(x)=3 x^2+2 x \mathrm{f}^{\prime}(1)+\mathrm{f}^{\prime \prime}(2)$, then $\mathrm{f}(x)=$ ______

Given the function $\mathrm{f}(x)=3x^2+2x\mathrm{f}^{\prime}(1)+\mathrm{f}^{\prime\prime}(2)$, the derivative terms are constants since they are evaluated at specific points.
Define $A = \mathrm{f}^{\prime}(1)$ and $B = \mathrm{f}^{\prime\prime}(2)$, thus $\mathrm{f}(x) = 3x^2 + 2A x + B$.

Differentiating yields $\mathrm{f}^{\prime}(x) = 6x + 2A$ and further, $\mathrm{f}^{\prime\prime}(x) = 6$.

From $\mathrm{f}^{\prime\prime}(x) = 6$, it follows that $B = 6$.

Substituting $x=1$ into $\mathrm{f}^{\prime}(x)$ gives $A = 6(1) + 2A$, leading to $A = -6$.

The function becomes $\mathrm{f}(x) = 3x^2 - 12x + 6$, which corresponds to option C.

$\boxed{\text{C}}$