jmc

From the second given congruence, we have $$c\equiv 3\cdot 5c\equiv 3\cdot 2\equiv 6(\mathrm{mod}7).$$From the third given congruence, we have $$5b\equiv 3(\mathrm{mod}7)$$$$⟹b\equiv 3\cdot 5b\equiv 3\cdot 3\equiv 2(\mathrm{mod}7).$$Then, from the first given congruence, we have $$1\equiv abca\cdot 6\cdot 2\equiv 12a\equiv 5a(\mathrm{mod}7)$$$$⟹a\equiv 3\cdot 5a\equiv 3\cdot 1\equiv 3(\mathrm{mod}7).$$Thus, $$a+b+c\equiv 3+2+6\equiv \boxed{4}(\mathrm{mod}7).$$