jmc

By AM-GM, $$\begin{array}{rl}x+y+z& =\frac{x}{3}+\frac{x}{3}+\frac{x}{3}+\frac{y}{2}+\frac{y}{2}+z\\ & \ge 6\sqrt[6]{\frac{x^3{y}^{2}z}{108}}.\end{array}$$Since $x+y+z=1,$ this gives us $$x^3{y}^{2}z\le \frac{108}{6^6}=\frac{1}{432}.$$Equality occurs when $\frac{x}{3}=\frac{y}{2}=z.$ Along with the condition $x+y+z=1,$ we can solve to get $x=\frac{1}{2},$ $y=\frac{1}{3},$ and $z=\frac{1}{6},$ so the maximum value is $\boxed{\frac{1}{432}}.$