原式=[(3a-7)+(a+5)]×[(3a-7)-(a+5)]÷[2(a-6)]
=[(4a-2)(2a-12)]÷[2(a-6)]
=[4(2a-1)(a-6)]÷[2(a-6)]
=2(2a-1)
当a=3时,原式=2(2×3-1)=10