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We rewrite the given equation, then rearrange: $$\begin{array}{rl}\frac{100a+10b+c}{999}& =\frac{1}{3}\left(\frac{a}{9}+\frac{b}{9}+\frac{c}{9}\right)\\ 100a+10b+c& =37a+37b+37c\\ 63a& =27b+36c\\ 7a& =3b+4c.\end{array}$$ Now, this problem is equivalent to counting the ordered triples $(a,b,c)$ that satisfies the equation. Clearly, the $9$ ordered triples $(a,b,c)=(1,1,1),(2,2,2),\dots ,(9,9,9)$ are solutions to this equation. The expression $3b+4c$ has the same value when: $b$ increases by $4$ as $c$ decreases by $3.$ $b$ decreases by $4$ as $c$ increases by $3.$ We find $4$ more solutions from the $9$ solutions above: $(a,b,c)=(4,8,1),(5,1,8),(5,9,2),(6,2,9).$ Note that all solutions are symmetric about $(a,b,c)=(5,5,5).$ Together, we have $9+4=\boxed{13}$ ordered triples $(a,b,c).$