Really finding roots

First of all, we have:

$(abc)^2=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=3$

hence $abc=a+b+c=\pm\sqrt3$ . Applying Vieta, we see that $a$ , $b$ and $c$ are the roots of the polynomial

$z^3-(a+b+c)z^2+(ab+bc+ca)z-(abc) = z^2\mp z^2\sqrt3+z\mp\sqrt3$

$=(z^2+1)(z\mp\sqrt3) =(z+i)(z-i)(z\mp\sqrt3).$

Therefore, $a,b,c$ are equal to $i,-i$ and either $\sqrt3$ or $-\sqrt3$ . In any case, we obtain:

$|a|^2+|b|^2+|c|^2=1^2+1^2+\sqrt3^2=\boxed{5}$ .