jmc

Let $q$ be the given quantity, $AMC+AM+MC+CA$. Notice that $$q+(A+M+C)+1=(A+1)(M+1)(C+1).$$By AM-GM, $$(A+1)(M+1)(C+1)\le {\left[\frac{(A+1)+(M+1)+(C+1)}{3}\right]}^3={\left(\frac{A+M+C+3}{3}\right)}^3=125,$$so $q\le 125-12-1=112.$

Equality occurs when $A=M=C=4,$ so the maximum value is $\boxed{112}.$