$\begin{aligned}
& \mathrm{S}=\left\{x \in \mathbb{R} / x^2+30 \leq 11 x\right\} \\
& x^2+30 \leq 11 x \\
& x^2-11 x+30 \leq 0 \\
& (x-5)(x-6) \leq 0 \\
& x \in[5,6]
\end{aligned}$
Now, $\mathrm{f}(x)=3 x^3-18 x^2+27 x-40$
$\begin{aligned}
\mathrm{f}^{\prime}(x) & =9 x^2-36 x+27 \\
\mathrm{f}^{\prime}(x) & =9\left(x^2-4 x+3\right) \\
& =9\left[\left(x^2-4 x+4\right)-1\right] \\
& =9(x-2)^2-9
\end{aligned}$
$\therefore \quad \mathrm{f}^{\prime}(x)>0 \forall x \in[5,6]$
$\therefore \mathrm{f}(x)$ is strictly increasing in the interval $[5,6]$
$\therefore \quad$ Maximum value of $\mathrm{f}(x)$ when $x \in[5,6]$ is
$f(6)=122$