We seek for large enough positive integers $a, b, c$ such that $\frac{ab+ac+bc}{a+b+c} = 3^N$ for some positive integer $N$. Indeed, one can write $\frac{ab+ac+bc}{a+b+c} = a+b-\frac{a^2+ab+b^2}{a+b+c}$. Now, choose $c$ such that $\frac{a^2+ab+b^2}{a+b+c} = 1$ that is, $c = a^2+b^2+ab-a-b$ then we need to find $a, b$ such that $a+b-1=3^N$. Choose coprime $a, b$ such that $3$ doesn't divide $ab(a-1)(b-1)$ it follows that $\text{gcd}(c, b) = \text{gcd}(c, a) = 1$. Further, choosing $a \equiv b \equiv 2 \pmod{3}$ implies that $3$ divides also $a+b+c = a^2+ab+b^2$. The rest is clear.