3. As shown in Figure 2, there is a point $O$ inside $\triangle A B C$. Lines parallel to each side are drawn through $O$, dividing $\triangle A B C$ into three triangles and three parallelograms. If the areas of the three triangles are $1,1,2$, then the area of $\triangle A B C$ is $\qquad$
3. $6+4 \sqrt{2}$.

As shown in Figure 7, it is easy to see that the three triangles are similar to $\triangle A B C$. Let the area of $\triangle A B C$ be $S$, then
$$
\begin{aligned}
& \frac{\sqrt{S_{1}}}{\sqrt{S}}+\frac{\sqrt{S_{2}}}{\sqrt{S}}+\frac{\sqrt{S_{3}}}{\sqrt{S}} \\
& =\frac{O R}{B C}+\frac{O T}{B C}+\frac{P Q}{B C} \\
& =\frac{B P+P Q+Q C}{B C}=1 . \\
\text { Therefore, } S & =\left(\sqrt{S_{1}}+\sqrt{S_{2}}+\sqrt{S_{3}}\right)^{2} \\
= & (1+1+\sqrt{2})^{2}=6+4 \sqrt{2} .
\end{aligned}
$$