2. As shown in Figure 3, select a point $P$ inside $\triangle A B C$, and draw line segments $D E, F G$, and $H M$ through $P$ parallel to the three sides of $\triangle A B C$. The areas of the resulting $\square A F P M, \square B D P G$, and $\square C E P H$ are $28, 12$, and $42$ respectively. Then the area of $\triangle A B C$ is $\qquad$
2.144.

Let the areas of $\triangle D P M$, $\triangle P G H$, and $\triangle P E F$ be $S_{1}$, $S_{2}$, and $S_{3}$, respectively. According to the problem,
$\triangle D P M \backsim \triangle P E F \backsim \triangle D E A$.
Therefore, $\frac{\sqrt{S_{1}}}{\sqrt{S_{\triangle A D E}}}=\frac{D P}{D E}$, $\frac{\sqrt{S_{3}}}{\sqrt{S_{\triangle D E E}}}=\frac{P E}{D E}$.
Thus, $\frac{\sqrt{S_{1}}+\sqrt{S_{3}}}{\sqrt{S_{\triangle A D E}}}=\frac{D P+P E}{D E}=\frac{D E}{D E}=1$, which means $\sqrt{S_{1}}+\sqrt{S_{3}}=\sqrt{S_{\triangle A D E}}$.
Since $S_{\square A F P M}=S_{\triangle A D E}-\left(S_{1}+S_{3}\right)$, we have $S_{1} S_{3}=\frac{1}{4} S_{\square A F P M}^{2}=\frac{1}{4} \times 28^{2}$.
Similarly, $S_{3} S_{2}=\frac{1}{4} S_{\square C E P H}^{2}=\frac{1}{4} \times 42^{2}$,
$S_{2} S_{1}=\frac{1}{4} S_{\square B D P G}^{2}=\frac{1}{4} \times 12^{2}$.
Thus, $S_{1}=4$, $S_{2}=9$, $S_{3}=49$.
Therefore, $S_{\triangle A B C}$
$$
\begin{array}{l}
=S_{\text {DAFPM }}+S_{\text {DBSPG }}+S_{\text {DCXPH }}+S_{1}+S_{2}+S_{3} \\
=28+12+42+4+9+49=144 .
\end{array}
$$