$\begin{array}{l}
\sqrt 3 \cot 3x = 1\\
\Leftrightarrow \cot 3x = - \dfrac{{\sqrt 3 }}{3}\\
\Leftrightarrow 3x = - \dfrac{\pi }{3} + k\pi \\
\Leftrightarrow x = - \dfrac{\pi }{9} + \dfrac{{k\pi }}{3}\left( {k \in \mathbb{Z}} \right)\\
k = 0 \Rightarrow x = - \dfrac{\pi }{9} \in \left( { - \dfrac{\pi }{2};0} \right)\\
k = - 1 \Rightarrow x = - \dfrac{{4\pi }}{9} \in \left( { - \dfrac{\pi }{2};0} \right)\\
k = 1 \Rightarrow x = \dfrac{{2\pi }}{9} \notin \left( { - \dfrac{\pi }{2};0} \right)\\
\Rightarrow S = - \dfrac{{5\pi }}{9} \to B
\end{array}$
