Find the Area

Let the line through $P$ parallel to $BC$ intersect $AB$ , $AC$ at $D$ , $E$ respectively. Again let the line through $P$ parallel to $CA$ intersect $BC$ , $AB$ at $F$ , $G$ respectively.Finally let the line through $P$ parallel to $AB$ intersect $BC$ , $CA$ at $K$ , $L$ respectively.

Assume that $\triangle PKF$ = $t_1$ , $\triangle PEL$ = $t_2$ and $\triangle PDF$ = $t_3$

Now, $\triangle PKF$ , $\triangle LPE$ , $\triangle GDP$ and $\triangle ABC$ are all similar triangle.

Also, $AGPL$ , $BDPK$ , $CEPF$ are all parallelograms

Next, $\frac{EC}{LE}$ = $\frac{PE}{LE}$ = $\sqrt{\frac{(KPL)}{(PLE)}}$ = $\sqrt{\frac{4}{9}}$ = $\frac{2}{3}$

Similarly, $\frac{AL}{LE}$ = $\frac{7}{3}$

So, $\frac{AC}{LE}$ = $\frac{AL+LE+EC}{LE}$ = $\frac{AL}{LE}$ + $\frac{LE}{LE}$ + $\frac{EC}{LE}$ = $4$

So, $\frac{(ABC)}{(LPE)}$ = $\frac{AC}{LE}^{2}$ = $16$

Which implies that $(ABC)$ = $144$

Three child triangles similar to their parent

Here all the three triangles are similar. So any triplet of their corresponding sides will be in a ratio of 2:3:7 (square roots of their areas)

Let us take their bases to be 2x, 3x, 7x which gives the base of the parent triangle to be (2x + 3x + 7x = 12x. So the area of the parent triangle must be $(2 + 3 + 7)^2 = \textbf {144}$