For non-zero real $x$, $y$, $z$ and complex $a$, $b$, $c$ with $|a|=|b|=|c|$, if $x+y+z=0=ax+by+cz$, then $a=b=c$

Assume that $\alpha = \beta \not= \gamma$.

We have $$(\gamma - \alpha)z =(\alpha)(x+y+z) + (\gamma - \alpha)z = \alpha x+ \beta y + \gamma z = 0$$

Since we assumed $\alpha \not= \gamma$, we have $z=0$, a contradiction.

Therefore, we have $\alpha = \beta = \gamma$, as desired. $\blacksquare$

$AB = 0$ does not exclude $A = B = 0$.

You can simply rearrange the equations:

$$ \begin{vmatrix} 1 & 1 & 1 \\ \cos\theta_2 & \cos\theta_3 & \cos\theta_1 \\ \sin\theta_2 & \sin\theta_3 & \sin\theta_1 \\ \end{vmatrix}=0$$

And use an analogous proof to show that ${\beta = \alpha} \vee {\beta = \gamma}$. And rearrange a final time to close the loop.

The only solution to $({\alpha = \beta} \vee {\alpha = \gamma}) \wedge ({\beta = \alpha} \vee {\beta = \gamma}) \wedge ({\gamma = \alpha} \vee {\gamma = \beta})$ is $\alpha = \beta = \gamma$.

Now give a different proof from above! First, we claim that: for any complex numbers $z_1,z_2$, we have $$|z_1+z_2|^2=|z_1|^2+|z_2|^2+2Re(z_1\overline{z_2}).$$ From
$$|\alpha|=|\beta|=|\gamma|=1$$ $$|x+y|^2=|-z|^2,$$ $$|\alpha x+\beta y|^2=|-\gamma z|^2.$$ Use above claim we can get: $$x^2+y^2+2xy=z^2,$$ $$x^2+y^2+2xyRe(\alpha\overline{\beta})=z^2,$$ so $Re(\alpha\overline{\beta})=1$ and $Im(\alpha\overline{\beta})=0$. (because $|\alpha|=|\beta|=1$)

We get $$\alpha=\beta.$$ In the same way, we can get $$\alpha=\beta=\gamma.$$