Answer: 108

Small Hint:

Use the two known cevian ratios to find the barycentric weights of $A$ and $B$ at $P$

Big Hint:

Place $P$ at the origin; the resulting vector relation determines the angle between $PA$ and $PB$

Solution:

Since $AP=PD,$ the barycentric weight of $A$ at $P$ is $\frac{1}{2}.$ Since $BP:PE=3:1,$ the weight of $B$ is $\frac{1}{4},$ so the weight of $C$ is also $\frac{1}{4}.$ Along $CF,$ this means $\frac{PF}{CF}=\frac{1}{4},$ hence $PF=5$ and $CP=15.$

Place $P$ at the origin and denote the position vectors $A,$ $B,$ and $C$ by the same letters. The barycentric relation is $2A+B+C=0.$ Thus $C=-2A-B.$ Using $|A|=6,$ $|B|=9,$ and $|C|=15,$ $$\begin{aligned}225&=|2A+B|^2\\&=4(36)+81+4A\mathbin{\cdot}B,\end{aligned}$$ so $A\mathbin{\cdot}B=0.$ Therefore $PA\perp PB.$ Expanding the cross product gives $(B-A)\mathbin{\times}(C-A)=4(A\mathbin{\times}B).$ Consequently, $$\begin{aligned}{}[ABC]&=2|A\mathbin{\times}B|\\&=2(6)(9)=108.\end{aligned}$$