96 Given $0 \leqslant \alpha \leqslant \beta<\gamma \leqslant 2 \pi$, and
$$
\cos \alpha+\cos \beta+\cos \gamma=0, \sin \alpha+\sin \beta+\sin \gamma=0
$$

then the value of $\beta-\alpha$ is
A. $\frac{\pi}{3}$
B. $\frac{2 \pi}{3}$
C. $\pi$
D. cannot be determined
96 B. Let the coordinates of points $A$, $B$, and $C$ be
$$
(\cos \alpha, \sin \alpha), (\cos \beta, \sin \beta) \text {, and } (\cos \gamma, \sin \gamma) \text {. }
$$

Then $\triangle A B C$ is an inscribed triangle in the circle $x^{2}+y^{2}=1$. Let the centroid $G$ of $\triangle A B C$ be $\left(x_{G}, y_{G}\right)$, then
$$
\begin{array}{l}
x_{G}=\frac{1}{3}(\cos \alpha+\cos \beta+\cos \gamma)=0, \\
y_{G}=\frac{1}{3}(\sin \alpha+\sin \beta+\sin \gamma)=0 .
\end{array}
$$

Thus, the centroid $G$ is also the circumcenter of $\triangle A B C$. Therefore, $\triangle A B C$ is an equilateral triangle, so $\beta-\alpha=\gamma-\beta=$ $\frac{2}{3} \pi$.