jmc

We start by changing the expressions to base 10 in terms of $a$ and $b$. We also know that the two expressions should be equal since they represent the same number. $$\begin{array}{rl}12_a& =21_b\Rightarrow \\ 1\cdot a+2\cdot 1& =2\cdot b+1\cdot 1\Rightarrow \\ a+2& =2b+1\Rightarrow \\ a& =2b-1.\end{array}$$For the smallest base-10 integer, we would want the smallest bases $a$ and $b$. Since $a$ and $b$ must be greater than 2, we'll let $b=3$ and that means $a=2\cdot 3-1=5$. In these bases, the base-10 integer is $a+2=5+2=\boxed{7}$. We can check that the base-$b$ expression also works and get $2\cdot b+1=2\cdot 3+1=7$.

Alternatively, we can just try different bases. The smallest possible value for $a$ and $b$ is 3. If we let $a=3$, we'd need a smaller base for $b$ (since we have $2\cdot b\approx 1\cdot a$), which isn't possible. When we let $b=3$, we get $21_3=7$ and try to find $b$ such that $12_b=7$. If $b+2=7$, then $b=5$ and we still get $\boxed{7}$.