∫ plane PAC⊥ plane ABC, plane PAC∩ plane ABC=AC,

∴BC⊥ Plane packaging,

∵PA？ Flat packaging, British Columbia, Pennsylvania ... (6 points)

(2) solution: let AD⊥PC be at point D. From the formula (1), we can know the bc⊥ flat packaging of BC⊥pc. ∴bc⊥ad.

PM∑BC and BC=2PM=4. The quadrilateral BCPM is a right-angled trapezoid with upper and lower bottoms of 2 and 4, height of 2 and area of 6 respectively.

BC∩PC=C, ∴AD⊥ plane BCPM, ad = 3.

Therefore, the volume of polyhedral PMBCA is13× sbcpm× ad =13× 6× 3 = 23 ... (13 minutes).