jmc

Every factor of $n$ is in the form $2^a\cdot 3^b\cdot 7^c$ for $0\le a\le 2$, $0\le b\le 1$, and $0\le c\le 2$. To count the number of even factors, we must restrict the power of 2 to be at least 1: $1\le a\le 2$. This gives us a total of $(2)(1+1)(2+1)=\boxed{12}$ even factors.