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[tex]3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{9\cdot12}=\sqrt{108}\\\\12\sqrt3=\sqrt{12^2\cdot3}=\sqrt{144\cdot3}=\sqrt{432}\\\\0,5\sqrt{32}=\sqrt{(0,5)^2\cdot32}=\sqrt{0,25\cdot32}=\sqrt8\\\\\sqrt8<\sqrt{108}<\sqrt{432}\\\\\huge\boxed{0,5\sqrt{32}<3\sqrt{12}<12\sqrt3}[/tex]
