The equation $3\phi(n)-n=c$ or $3\phi(n)-2n=c$ has only finitely many solutions for any fixed $c \ge 1.$

Question

A problem I'm trying to solve needs this result to proceed forward. I suspect that even though $\phi(n)/n$ gets arbitrarily close to $1/3,$ the result is true. I've shown that we can write $n = ab$ with $a$ bounded and $b$ squarefree. This implies that as $n \to \infty, n$ has arbitrarily many prime factors. Now let $n = \prod\limits_i p_i^{e_i}, e_i \ge 1.$ Then $\phi(n)/n = \prod\limits_i (1-\frac{1}{p_i}),$ so we require $\left|\prod\limits_i (1-\frac{1}{p_i}) - \frac{1}{3} \right| = \frac{c}{3n} \le \frac{c}{3} \prod\limits_i \frac{1}{p_i}$ for the first equation. As the primes grow large enough, there shouldn't be an approximation of $1/3$ that gets tremendously better (by at least a factor of $p$) as we tack on an extra term of $1-\frac{1}{p}$ to the product. Rigorously justifying this would imply the result and in fact allow us to tackle the more general equation $a\phi(n)-bn = c$ (just replace $1/3, c/3$ with $b/a, c/a$), but how do we do that?