Given, $$\omega \ne 1$$ be a cube root of unity, then $${\left| {a + b\omega + c{\omega ^2}} \right|^2}$$

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

[$$ \because $$ $$z\overline z = |z{|^2}$$]

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

[$$ \because $$ $${\overline \omega }$$ = $$\omega $$² and $${{{\overline \omega }^2}}$$ = $$\omega $$]

= $${a^2} + {b^2} + {c^2} + ab({\omega ^2} + \omega ) + bc({\omega ^2} + {\omega ^4}) + ac(\omega + {\omega ^2})$$

[as $${\omega ^3} = 1$$]

$$ = {a^2} + {b^2} + {c^2} + ab( - 1) + bc( - 1) + ac( - 1)$$

[as $$\omega + {\omega ^2} = - 1,\,{\omega ^4} = \omega $$]


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

$$ = {1 \over 2}\{ {(a - b)^2} + {(b - c)^2} + {(c - a)^2}\} $$

$$ \because $$ a, b and c are distinct non-zero integers. For minimum value a= 1, b = 2 and c = 3

$$ \therefore $$ $$|a + b\omega + c{\omega ^2}|_{\min }^2 = {1 \over 2}\{ {1^2} + {1^2} + {2^2}\} = {6 \over 2} = 3.00$$