6. In the planar quadrilateral $ABCD$, $AC$ intersects $BD$ at point $O$. If
$$
\begin{array}{l}
\overrightarrow{AC} + 2 \overrightarrow{DO} = 4 \overrightarrow{AB}, \\
3 AB = 2 CD, \\
\overrightarrow{BA} \cdot \overrightarrow{BD} = 9,
\end{array}
$$

then the maximum value of the area of $\triangle ACD$ is $\qquad$.
6. $4 \sqrt{3}$.

Extend $A B$ to point $E$, such that $A E=4 A B$, and take point $F$ on line $C E$ such that $B F / / A C$.
From $\overrightarrow{A C}+2 \overrightarrow{D O}=4 \overrightarrow{A B}$
$\Rightarrow 2 \overrightarrow{D O}=4 \overrightarrow{A B}-\overrightarrow{A C}=\overrightarrow{C E}$
$\Rightarrow D O / / C E$
$\Rightarrow$ Quadrilateral $B O C F$ is a parallelogram
$\Rightarrow D O=\frac{1}{2} C E=2 O B$
$\Rightarrow S_{\triangle A C D}=4 S_{\triangle A O D}=8 S_{\triangle A O B}$.
Let $\angle A B D=\theta$. Then
$$
\begin{array}{l}
S_{\triangle A C D}=4 S_{\triangle A O D}=8 S_{\triangle A O B} \\
=4 A B \cdot B O \sin \theta . \\
\text { Also } \overrightarrow{B A} \cdot \overrightarrow{B D}=9 \Rightarrow B A \cdot B O \cos \theta=3 .
\end{array}
$$

Thus $S_{\triangle A C D}=12 \tan \theta$.
From $3 A B=2 C D \Rightarrow D F=2 C D$.
Also, $\angle D F C=\theta \Rightarrow \tan \theta \leqslant \frac{1}{\sqrt{3}}$
$$
\Rightarrow S_{\triangle A C D} \leqslant 4 \sqrt{3} \text {. }
$$