smc

We can rewrite the given equation as $2^{a_7-27}=a_7$. Hence, $a_7$ must be a power of $2$ and larger than $27$. The first power of 2 that is larger than $27$, namely $32$, does satisfy the equation: $2^{32-27}=2^5=32$. In fact, this is the only solution; $2^{a_7-27}$ is exponential whereas $a_7$ is linear, so their graphs will not intersect again. Now, let the common difference in the sequence be $d$. Hence, $a_0=32-7d$ and $a_2=32-5d$. To minimize $a_2$, we maxmimize $d$. Since the sequence contains only positive integers, $32-7d>0$ and hence $d\le 4$. When $d=4$, $a_2=\boxed{12}$.