1. Given function and domain

We need the maximum of $f(x)=3x^3-18x^2+27x-40$ on the set $S=\{x\in \mathbb R: x^2+30\le 11x\}.$

So first find the allowed values of $x$.

2. Solve the inequality for $S$

$x^2+30\le 11x$ $x^2-11x+30\le 0$ $(x-5)(x-6)\le 0$

Hence, $S=[5,6].$

So we must maximize $f(x)$ on the closed interval $[5,6]$.

3. Find critical points

Differentiate: $f'(x)=9x^2-36x+27$ $f'(x)=9(x^2-4x+3)=9(x-1)(x-3).$

Critical points are $x=1,3$, but neither lies in $[5,6]$.

4. Check monotonicity on $[5,6]$

For $x\in[5,6]$, both $(x-1)>0$ and $(x-3)>0$, so $f'(x)=9(x-1)(x-3)>0.$

Therefore, $f(x)$ is strictly increasing on $[5,6]$.

Hence the maximum value occurs at the right endpoint $x=6$.

5. Evaluate at endpoints

At $x=5$: $f(5)=3(125)-18(25)+27(5)-40$ $=375-450+135-40=20.$

At $x=6$: $f(6)=3(216)-18(36)+27(6)-40$ $=648-648+162-40=122.$

Thus, $\max_{x\in S} f(x)=122.$

6. Compare with options

The correct option is: $\boxed{\text{C: }122}$

7. Verification with stored answer

Stored correct answer is C, which matches our result.