The prove I tried is the following. I really wish someone can check if I made some logical mistake, especially the last part I found myself diffident proving $a$ can only be $1$ or $5$.
Because $b\neq c$ and $b<c$, otherwise switch the value of $b,c$. Becuase $a\mid (3b+2c)$ and $a \mid (3c+2b), \exists q_1,q_2 \in N$ that $$aq_1=(3b+2c), aq_2=(3c+2b), a(q_2-q_1)=c-b.$$ Therefore we have $a\mid (c-b)$ and $a \mid 3(c-b)+(3b+2c)$ which is $a \mid 5c$. And we have $a \mid 2(c-b)+(3c+2b)$, $a\mid 5c$. Hence $a \mid 5b$ and $a \mid 5c$. Because $b,c$ are distinct odd primes. This is possible only when $a\mid b$ and $a \mid c$ or $a \mid 5$. The only number that divides two primes is $1$. And the two numbers that divides $5$ are $1$ and $5$. Hence the two possible values for $a$ are $1$ and $5$.