jmc

Let $a$ be a common term. We know that $$\begin{array}{rl}a& \equiv 2(\mathrm{mod}3)\\ a& \equiv 3(\mathrm{mod}7)\end{array}$$ Congruence $(1)$ means that there exists a non-negative integer such that $a=2+3n$. Substituting this into $(2)$ yields $$2+3n\equiv 3(\mathrm{mod}7)⟹n\equiv 5(\mathrm{mod}7)$$ So $n$ has a lower bound of $5$. Then $n\ge 5⟹a=2+3n\ge 17$. $17$ satisfies the original congruences, so it is the smallest common term. Subtracting $17$ from both sides of both congruences gives $$\begin{array}{rl}a-17& \equiv -15\equiv 0(\mathrm{mod}3)\\ a-17& \equiv -14\equiv 0(\mathrm{mod}7)\end{array}$$ Since $\gcd (3,7)$, we get $a-17\equiv 0(\mathrm{mod}3\cdot 7)$, that is, $a\equiv 17(\mathrm{mod}21)$.

So all common terms must be of the form $17+21m$ for some non-negative integer $m$. Note that any number of the form also satisfies the original congruences. The largest such number less than $500$ is $17+21\cdot 22=\boxed{479}$.