3. As shown in Figure 2, fold $\triangle A B C$ along the dotted line $D E$ to get a heptagon $A D E C F G H$. If the area ratio of the heptagon to the original triangle is $2: 3$, and the area of the overlapping part after folding is 4, then the area of the original $\triangle A B C$ is
3. 12 .

Let the area of the non-overlapping part after folding be $x$. Then the area of the original triangle is $8+x$, and the area of the heptagon is $4+x$. From the given condition, we have
$$
\begin{array}{l}
(8+x):(4+x)=3: 2 \\
\Rightarrow 16+2 x=12+3 x \Rightarrow x=4 .
\end{array}
$$
$$
\text { Hence } S_{\triangle A B C}=12 \text {. }
$$