An algebra problem by Suparno Karmakar

We write the given sum as $1 = a+2b+3c = (\tfrac13a)+ (\tfrac13a)+ (\tfrac13a)+b+b+(3c).$ With AM-GM, $\frac{(\tfrac13a)+ (\tfrac13a)+ (\tfrac13a)+b+b+(3c)}6 \geq \sqrt[6]{(\tfrac13a)\cdot(\tfrac13a)\cdot (\tfrac13a)\cdot b\cdot b\cdot (3c)}.$ $\frac16 \geq \sqrt[6]{\frac19a^3b^2c}.$ Equality holds when $\tfrac13a = b = 3c$ ; for that case, $a^3b^2c = 9\cdot \left(\frac16\right)^6 = \boxed{\frac1{5184}}.$

The problem should consider all a,b,c are positive or all negative. If don't, the question would be no answer. Because, for example, let a and c go to infinitive negative then for a + 2b +3c =1, b should go to infinity, then a^3 b^2 c go to infinity no maxium.