GCD and maximum

Let $d = \gcd\left(x, y, z\right)$ .

Then we may write $x = da, y = db, y = dc$ for some integers $a, b, c$ satisfying $\gcd\left(a, b, c\right) = 1$ .

With this representation, the given equation simplifies to $\gcd\left(a + 3b, b + 3c, c + 3a\right)=n$ .

This implies $n$ must also divide $\left(a + 3b\right) + 3\left(b + 3c\right) + 9\left(c + 3a\right) = 28a + 6(b + 3c)$ .

Then because $n$ divides $b + 3c$ , the above equation implies that $n$ also divides $28a$ .

Similar reasoning shows that $n$ divides $28b$ and $28c$ .

It follows that $n$ divides $\gcd\left(28a, 28b, 28c\right) = 28\cdot \gcd\left(a, b, c\right) = 28$ .

Hence $n\le 28$ . We can show that $n = 28$ is achievable by taking $\left(x, y, z\right) = \left(13, 5, 17\right)$ .

Indeed, in this case $\gcd\left(x, y, z\right) = 1$ and $\gcd\left(x + 3y, y + 3z, z + 3x\right) = \gcd\left(28, 56, 56\right) = 28$ .

So $\boxed{n = 28}$ is the answer.