jmc

Note that $|{\omega }^3|=|\omega {|}^3=1,$ so $|\omega |=1.$ Then $\omega \overline{\omega }=|\omega {|}^2=1.$

Also, ${\omega }^3-1=0,$ which factors as $(\omega -1)({\omega }^2+\omega +1)=0.$ Since $\omega \ne 1,$ $${\omega }^2+\omega +1=0.$$Hence, $$\begin{array}{rl}|a+b\omega +c{\omega }^2{|}^2& =(a+b\omega +c{\omega }^2)(a+b\overline{\omega }+c\overline{{\omega }^2})\\ & =(a+b\omega +c{\omega }^2)\left(a+\frac{b}{\omega }+\frac{c}{{\omega }^2}\right)\\ & =(a+b\omega +c{\omega }^2)(a+b{\omega }^2+c\omega )\\ & =a^2+b^2+c^2+(\omega +{\omega }^2)ab+(\omega +{\omega }^2)ac+({\omega }^2+{\omega }^4)bc\\ & =a^2+b^2+c^2+(\omega +{\omega }^2)ab+(\omega +{\omega }^2)ac+(\omega +{\omega }^2)bc\\ & =a^2+b^2+c^2-ab-ac-bc\\ & =\frac{(a-b{)}^2+(a-c{)}^2+(b-c{)}^2}{2}.\end{array}$$Since $a,$ $b,$ and $c$ are distinct, all three of $|a-b|,$ $|a-c|,$ and $|b-c|$ must be at least 1, and at least one of these absolute values must be at least 2, so $$\frac{(a-b{)}^2+(a-c{)}^2+(b-c{)}^2}{2}\ge \frac{1+1+4}{2}=3.$$Equality occurs when $a,$ $b,$ and $c$ are any three consecutive integers, in any order, so the smallest possible value of $|a+b\omega +c{\omega }^2|$ is $\boxed{\sqrt{3}}.$