If a,b and c are distinct integers and $\omega (\ne 1)$ is a cube root of unity, then the minimum value of $|a+b\omega +c{\omega }^2|+|a+b{\omega }^2+c\omega |,is$

$Letz=a+b\omega +c{\omega }^2.$ then

${\left|z\right|}^2=z\stackrel{-}{z}=(a+b\omega +c{\omega }^2)(a+b\stackrel{-}{\omega }+c\stackrel{-}{{\omega }^2})$

$=(a+b\omega +c{\omega }^2)(a+b\stackrel{ }{{\omega }^2}+c\stackrel{ }{{\omega }^{ }})$

$= a^2+b^2+c^2-ab-bc-ca$

$= \frac{1}{2}[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]$

$\Rightarrow {\left|z\right|}^2\ge \frac{1}{2}\times 6=3\left[\begin{array}{l}\therefore a\ne b\ne c\\ \therefore |a-b|\ge 1,|b-c|\ge 1\\ and|a-c|\ge 2\end{array}\right]$

$\therefore \left|z\right|\ge \sqrt{3}$

$\therefore |a+b\omega +c{\omega }^2|+|a+b{\omega }^2+c\omega |$

$=|a+b\omega +c{\omega }^2|+|a+b\stackrel{-}{{\omega }^2}+c\stackrel{-}{\omega }|$

$= |a+b\omega +c{\omega }^2|+|a+b\omega +c\stackrel{ }{{\omega }^2}|$

$2|a+b\omega +c{\omega }^2|=2\left|z\right|\ge 2\sqrt{3}$

Hence, the minimum value of $|a+b\omega +c{\omega }^2|+|a+b{\omega }^2+c\omega |is2\sqrt{3}$