【Solution】Solution: According to the analysis, $S_{\triangle B D C}=S_{\triangle E B C} \Rightarrow S_{\triangle D O B}=S_{\triangle E O C}$,
$$
\therefore S_{\text {甲 }}-S_{\text {乙 }}=\left(S_{\text {甲 }}+S_{\triangle D O B}\right)-\left(S_{\text {乙 }}+S_{\triangle E O C}\right)=5.04,
$$

Also, $\because S_{\triangle B D C}: S_{\triangle D E C}=B C: D E=2: 1$ that is: $S_{\triangle B D C}=2 S_{\triangle D E C}$
$\therefore S_{\text {quadrilateral } D E C B}=3 S_{\triangle D E C} ; S_{\triangle A D E}=S_{\triangle D E C}$
$\therefore S_{\triangle A B C}=S_{\text {quadrilateral } D E C B}+S_{\triangle A D E}=4 S_{\triangle D E C}$,
Let $S_{\triangle D E C}=X$, then $S_{\triangle B D C}=2 X$, so we have $2 X-X=5.04$,
$$
\therefore X=5.04, S_{\triangle A B C}=4 S_{\triangle D E C}=4 X=4 \times 5.04=20.16
$$

Therefore, the answer is: 20.16
12. (6 points) As shown in the figure, in $\triangle A B C$, $D$ and $E$ are the midpoints of $A B$ and $A C$, respectively, and the area difference between the two shaded regions (甲 and 乙) is 5.04. Then $S_{\triangle A B C}=$ $\qquad$ .