If PB is extended to D, PD = 2PB, PC is extended to E, PE=3PC, DF‖PE, EF‖PD, then PDFE is a parallelogram, DE and PF intersect at point O, and PO=OF.

Vector (PD+PE)=2 Vector PB+3 Vector PC= Vector PF=2 Vector PO=- Vector PA

∴A,P,O * * line, PO=AP/2.

S△BCP = S△OCP = S△ACP/2 = = = = & gt； S△ACP=2S△BCP

S△BCP=S△OCP=S△OEP/3，S△OBP = S△OEC/2 = = = & gt； S△OBP= 1.5S△BCP

S△ABP=2S△OBP=3S△BCP