The given function is $f\left(x\right)=(x-3{)}^{2018}(x-2{)}^{2019}, x\in \mathrm{ℝ}$.

Differentiating w.r.t. $x$ we get, 

$f^{\text{'}}\left(x\right)={\left(x-3\right)}^{2017}{\left(x-2\right)}^{2018}\left\{2018\left(x-2\right)+2019\left(x-3\right)\right\}$

$\Rightarrow f^{\text{'}}\left(x\right)={\left(x-3\right)}^{2017}4037{\left(x-2\right)}^{2018}\left\{x-\frac{10093}{4037}\right\}$

Using rate of change of $f^{\text{'}}\left(x\right)$, we can conclude that there is a local maxima at $x=\frac{10093}{4037}$.

$\Rightarrow \alpha =\frac{10093}{4037}$.

Therefore, $2\alpha +3f\left(\alpha \right)=\frac{20186}{4037}-3{\left(\frac{2018}{4037}\right)}^{2018}{\left(\frac{2019}{4037}\right)}^{2019}$.