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Let Amy's, Ben's, and Chris's ages be $a$, $b$, and $c$, respectively. We have the equations $$\begin{array}{rl}\frac{a+b+c}{3}=9\Rightarrow a+b+c& =27\\ c-4& =a\\ b+3& =\frac{2}{3}(a+3)\end{array}\text{\hspace{1em}(1)(2)(3)}$$ From Equation (3), we have $b=\frac{2}{3}(a+3)-3$. We substitute Equation (2) into Equation (3) to eliminate $a$, to get $b=\frac{2}{3}(c-1)-3$. Substituting this last equation and Equation (2) into Equation (1) to eliminate $a$ and $b$, we have $$[c-4]+[\frac{2}{3}(c-1)-3]+c=27$$ Solving for $c$, we find that $c=13$. Thus, Chris's age is $\boxed{13}$.