IIT-JEE 2011 · Paper 2, Mathematics question 53

Official key: marks to all.

Why. $\omega=e^{i\pi/3}$ is a sixth root of unity, not a cube root, so $1+\omega+\omega^2=2\omega\ne0$ and the ratio depends on $a$, $b$, $c$.

Intended reading ($\omega=e^{2\pi i/3}$, so $1+\omega+\omega^2=0$):
$$|x|^2+|y|^2+|z|^2=3\left(|a|^2+|b|^2+|c|^2\right),$$
because all cross terms carry a factor $1+\omega+\omega^2$. That gives the answer 3.