Question

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Let \(D,E,F\) be points on the sides \(BC,CA, AB\) respectively of a triangle \(ABC\). The lines \(AD,BE, CF\) are concurrent at the point \(P\). If \(AP = PD= 2\) , \(BP = 3\), \(PE=1\) , \(PF=\frac {5}{3}\) and \(CP = 5\), then area of triangle?