jmc

By Cauchy-Schwarz, $$(4\sqrt{a}+6\sqrt{b}+12\sqrt{c}{)}^2\le (4^2+6^2+12^2)(a+b+c)=(196)(4abc)=784abc,$$so $$4\sqrt{a}+6\sqrt{b}+12\sqrt{c}\le 28\sqrt{abc},$$and $$\frac{4\sqrt{a}+6\sqrt{b}+12\sqrt{c}}{\sqrt{abc}}\le 28.$$Equality occurs when $$\frac{a}{16}=\frac{b}{36}=\frac{c}{144}.$$Along with the condition $a+b+c=4abc,$ we can solve to get $a=\frac{7}{18},$ $b=\frac{7}{8},$ $c=\frac{7}{2}.$ Therefore, the maximum value is $\boxed{28}.$