Three, through an internal point $Q$ of $\triangle A B C$, draw lines not parallel to the three sides of the triangle. These lines divide the triangle into six parts. It is known that the areas of the three parallelogram parts are $S_{1}, S_{2}, S_{3}$ (as shown in the figure), find the area of $\triangle A B C$.

Draw lines not parallel to the three sides of the triangle through an internal point $Q$ of $\triangle A B C$. These lines divide the triangle into six parts. It is known that the areas of the three parallelogram parts are $S_{1}, S_{2}, S_{3}$ (as shown in the figure), find the area of $\triangle A B C$.
Three, Solution: For convenience, let $S: q \nu G, S: Q$: $S \triangle Q P \mathrm{Rg}$ be $S_{1}^{\prime}, S_{2}^{\prime}, S_{3}^{\prime}$. Since
$$
\frac{\frac{1}{2} S_{1}}{S_{2}^{\prime}}=\frac{S \Delta Q B E}{S_{2}^{\prime}}=\frac{B E}{E I}=\frac{F Q}{E I}=\frac{H Q}{Q I}=\frac{S_{2}}{S_{s}},
$$

we have $S_{2}^{\prime}=\frac{S_{3} S_{1}}{2 S_{2}}$.
Similarly, $S_{1}^{\prime}=\frac{S_{2} S_{3}}{2 S_{1}}, \quad S_{8}^{\prime}=\frac{S_{1} S_{2}}{2 S_{9}}$.
Therefore, $S_{\triangle A B C}=S_{1}+S_{2}+S_{3}$
$$
+\frac{S_{2} S_{3}}{2 S_{1}}+\frac{S_{8} S_{1}}{2 S_{2}}+\frac{S_{1} S_{2}}{2 S_{3}} \cdot
$$