$4 \cdot 180$ Try to determine all the quadruples $\left(p_{1}, p_{2}, p_{3}, p_{4}\right)$, where $p_{1}, p_{2}$, $p_{3}, p_{4}$ are all prime numbers, and satisfy:
(1) $p_{1}<p_{2}<p_{3}<p_{4}$;
(2) $p_{1} p_{2}+p_{2} p_{3}+p_{3} p_{4}+p_{4} p_{1}=882$.