In the interior of triangle $ABC$, we have chosen point $P$ such that the lines drawn through $P$ parallel to the sides of the triangle determine 3 triangles and 3 parallelograms. The areas of the resulting triangles are 4, 9, and 49 square units. What is the area of the original triangle?
We denote the sides of the small triangles parallel to $AB = c$ as $c_1$, $c_2$, and $c_3$ respectively. The opposite sides of the parallelograms are equal, so $c = c_1 + c_2 + c_3$. Due to the parallelism, the small triangles are similar to the triangle $ABC$, and the ratio of their linear dimensions is given by the ratio of the parallel sides $\frac{c_1}{c}$, $\frac{c_2}{c}$, and $\frac{c_3}{c}$.

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It is known that the ratio of the areas of similar figures is the square of the ratio of their linear dimensions. If we denote the area of triangle $ABC$ by $T$, and the areas of the corresponding small triangles by $T_1$, $T_2$, and $T_3$, then from the above, we have

$$
\frac{T_1}{T} = \frac{c_1^2}{c^2}, \quad \frac{T_2}{T} = \frac{c_2^2}{c^2}, \quad \frac{T_3}{T} = \frac{c_3^2}{c^2}
$$

Taking the square root of both sides of these three equations and then adding the resulting equations, we get

$$
\frac{\sqrt{T_1} + \sqrt{T_2} + \sqrt{T_3}}{\sqrt{T}} = \frac{c_1 + c_2 + c_3}{c} = 1
$$

from which

$$
T = \left(\sqrt{T_1} + \sqrt{T_2} + \sqrt{T_3}\right)^2
$$

The area of triangle $ABC$ is thus

$$
T = (2 + 3 + 7)^2 = 144 \text{ square units. }
$$