$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}4\mathrm{x}-{\mathrm{x}}^3+\ln \left({\mathrm{a}}^2-3\mathrm{a}+3\right),\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\le \mathrm{x}<3\\ \mathrm{x}-18,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\ge 3\end{array}\right.$
Complete set of values of $a$ such that $f\left(x\right)$ as a local minima at  $x=3$ is

Clearly, $f\left(x\right)$ in decreasing just before $x=3$ and increasing after $x=3$. For $x=3$ to be the point of local minima,$\mathrm{f}\left(3\right)\le \mathrm{f}\left(3^{-}\right)$.
or    $-15\le 12-27+\ln \left({\mathrm{a}}^2-3\mathrm{a}+3\right)$
or   ${\mathrm{a}}^2-3\mathrm{a}+3\ge 1$
or   $\mathrm{a}\in (-\mathrm{\infty },1)\cup (2,\mathrm{\infty })$