6. As shown in Figure 1, in the right trapezoid $A B C D$, $\angle B=\angle C$ $=90^{\circ}, \angle A M C=120^{\circ}$, $A M=B C=6, C D=\sqrt{3}$. Let the circumcenters of $\triangle M C D$, $\triangle D M A$, and $\triangle A B M$ be $O_{1}$, $O_{2}$, and $O_{3}$, respectively. Then the area of $\triangle O_{1} O_{2} O_{3}$ is ( ).
(A) $\frac{3 \sqrt{3}}{2}$
(B) $2 \sqrt{3}$
(C) $3 \sqrt{3}$
(D) $4 \sqrt{3}$
6.A.

As shown in Figure 5,
$$
\begin{array}{l}
\angle A M C=120^{\circ} \\
\Rightarrow \angle A M B=60^{\circ}
\end{array}
$$
$\Rightarrow$ In the right triangle $\triangle A B M$,
$$
\begin{aligned}
& \angle M A B=30^{\circ} \\
\Rightarrow & M B=3=M C \\
\Rightarrow & D M=2 \sqrt{3}
\end{aligned}
$$
$\Rightarrow$ In the right triangle $\triangle D C M$,
$$
\angle D M C=30^{\circ}
$$
$\Rightarrow \triangle D M A$ is also a right triangle
$$
\Rightarrow A D=4 \sqrt{3} \text {. }
$$

Since the centers $O_{1}, O_{2}, O_{3}$ are the midpoints of $D M, D A, A M$ respectively, by the midline theorem we have
$$
\begin{array}{l}
O_{1} O_{2}=\frac{1}{2} A M=3, O_{2} O_{3}=\frac{1}{2} D M=\sqrt{3}, \\
O_{1} O_{3}=\frac{1}{2} A D=2 \sqrt{3} .
\end{array}
$$

Thus, $\triangle O_{1} O_{2} O_{3}$ is also a right triangle.
Hence, $S_{\triangle O_{1} O_{2} O_{3}}=\frac{3 \sqrt{3}}{2}$.