3. As shown in the figure, $P$ is a point inside $\triangle ABC$. Lines are drawn through $P$ parallel to the sides of $\triangle ABC$.
The smaller triangles
$t_{1}, t_{2}$, and $t_{3}$
have areas of 4, 9,
and 49, respectively. Find the area of $\triangle ABC$.
Given the figure, let $\mathrm{MP}=\mathrm{p}, \mathrm{PN}=\mathrm{q}, \mathrm{RT}=\mathrm{r}$. Since triangles $t_{1}, t_{2}, t_{3}$ are all similar to $\triangle A B C$, the ratio of their areas is equal to the square of the ratio of corresponding sides.
Let $\mathrm{AB}=c, \triangle \mathrm{ABC}$ have an area of $\mathrm{S}$, then we have:
$$
\sqrt{2}=\frac{q}{c}, \frac{3}{\sqrt{S}}=\frac{p}{c}, \frac{7}{\sqrt{S}}=\frac{r}{c}
$$

Adding the three equations, we get: $\frac{2+3+7}{\sqrt{\bar{S}}}=\frac{\mathrm{p}+\mathrm{q}+\mathrm{r}}{\mathrm{c}}$
$$
\begin{aligned}
=\frac{c}{c} & =1 \\
\therefore \sqrt{S} & =12, S=144
\end{aligned}
$$