2. Let $z_{1}=1+i, z_{2}=2+i, z_{3}=3+i$. Then $\arg \left(-10 \sqrt{3}+z_{1} z_{2} z_{3}\right)$ equals ( ).
(A) $\frac{\pi}{6}$
(B) $\frac{\pi}{3}$
(C) $\frac{2 \pi}{3}$
(D) $\frac{5 \pi}{6}$
2. (D)
$$
\begin{array}{l}
\text { Since } z_{1} z_{2} z_{3}=(1+i)(2+i)(3+i)=10 i, \\
\therefore \arg \left(-10 \sqrt{3}+z_{1} z_{2} z_{3}\right) \\
=\arg (-10 \sqrt{3}+10 i)=\frac{5 \pi}{6} .
\end{array}
$$