jmc

We apply Simon's Favorite Factoring Trick and note that if we subtract 12 from both sides, then the left side can be factored. Thus, $$ab-3a+4b-12=125\to (a+4)(b-3)=125$$Since $a,b$ are positive integers, then $a+4,b-3$ must be a pair of factors of $125=5^3$, so $(a+4,b-3)$ must be among $$(1,125),(5,25),(25,5),(125,1).$$Thus, $(a,b)$ must be among $$(-3,128),(1,28),(21,8),(121,4).$$Ruling out the first solution on account of the negative value for $a$, we find that the minimal value of $|a-b|$ among the remaining three is $|21-8|=\boxed{13}$.