Three, the maximum possible value of $n$ is 9.
First, prove that $a+b+c$ can be divided by 3.
In fact, $a+b+c=a+b+2a+5b=3(a+2b)$, so $a+b+c$ is a multiple of 3.
Let the remainders when $a$ and $b$ are divided by 3 be $r_{a}$ and $r_{b}$, respectively, then $r_{a} \neq 0, r_{b} \neq 0$.
If $r_{a} \neq r_{b}$, then $r_{a}=1, r_{b}=2$ or $r_{a}=2, r_{b}=1$. In this case, $2a+5b$ must be a multiple of 3, i.e., $c$ is a composite number, which is a contradiction.
Therefore, $r_{a}=r_{b}$, then $r_{a}=r_{b}=1$ or $r_{a}=r_{b}=2$,
In this case, $a+2b$ must be a multiple of 3, thus $a+b+c$ is a multiple of 9.
Next, prove that 9 is the largest.
$$\because 2 \times 11+5 \times 5=47 \text {, and } 11+5+47=63 \text {, }$$

While in $2 \times 13+5 \times 7=61$, $13+7+61=81$, and $(63, 81)=9$, hence 9 is the largest possible value.