\[\begin{align}
& {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc \\
& {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0 \\
& {{(a+b)}^{3}}+{{c}^{3}}-3ab(a+b)-3abc=0 \\
& {{(a+b+c)}^{3}}-3(a+b)c(a+b+c)-3ab(a+b+c)=0 \\
& (a+b+c)\left[ {{(a+b+c)}^{2}}-3ab-3ac-3bc \right]=0 \\
& (a+b+c)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)=0 \\
& \frac{1}{2}(a+b+c)\left[ {{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}} \right]=0 \\
& \Rightarrow a=b=c=\frac{1}{3} \\
& Q=5a+6b+2019c=\frac{2030}{3} \\
\end{align}\]