$n=\frac{abc+ab+a}{abc+bc+c}$ $abc+ab+a \leq 3abc<3(abc+bc+c)$ so $n<3$ $n=1 \to a=c$ For $n=2$ $abc+ab+a=2(abc+bc+c)$ $abc+2bc+2c-ab-a=0$ so $abc<ab+a \to bc<b+1 \to c=1$ $2b+2=a$ and checking $\frac{abc+ab+a}{abc+bc+c}=\frac{(2b+2)b+(2b+2)b+(2b+2)}{(2b+2)b+b+1}=2$