3.74 Determine all quadruples $\left(p_{1}, p_{2}, p_{3}, p_{4}\right)$, where $p_{1}, p_{2}, p_{3}$, $p_{4}$ are prime numbers, and satisfy
(1) $p_{1}<p_{2}<p_{3}<p_{4}$;
(2) $p_{1} p_{2}+p_{2} p_{3}+p_{3} p_{4}+p_{4} p_{1}=882$.
(Australian Mathematical Olympiad, 1995)
[Solution] Condition (2) can be transformed into $\left(p_{1}+p_{3}\right)\left(p_{2}+p_{4}\right)=882=2 \times 3^{2} \times 7^{2}$.
Since $4 \nmid 882$, one of $p_{1}+p_{3}$ and $p_{2}+p_{4}$ must be odd. Also, since $p_{1}$, $p_{2}$, $p_{3}$, $p_{4}$ are prime numbers and $p_{1}$ is the smallest, we have $p_{1}=2$, thus $2+p_{3}$ is an odd divisor of 882.
That is,
$$
2+p_{3} \mid 441 \text {. }
$$

Since $p_{1}+p_{3}<p_{2}+p_{4}$,
we have $2+p_{3}<\sqrt{882}<30$.
Therefore, the possible values of $p_{3}+2$ can only be $1,3,7,9,21$. Given that $p_{3}$ is a prime number, $p_{3}$ can only be $5,7,19$.
We will discuss each case separately,
(i) If $p_{3}=5$, then by condition (1), $p_{2}=3$, and $p_{2}+p_{4}$ should be $\frac{882}{2+5}=126$, so $p_{4}=123$ is not a prime number;
(ii) If $p_{3}=7$, then $p_{2}+p_{4}=\frac{882}{2+7}=98$, and $p_{2}$ can only be 3 or 5, corresponding to $p_{4}$ being 95 or 93, neither of which are prime numbers.
(iii) If $p_{3}=19$, then $p_{2}+p_{4}=\frac{882}{2+19}=42$,
Since
$$
2<p_{2}<19<p_{4},
$$

then $p_{2}=5,11,13$, and correspondingly, $p_{4}=37,31,29$.
In summary, the quadruples $\left(p_{1}, p_{2}, p_{3}, p_{4}\right)$ are $(2,5,19,37),(2,11,19,31),(2,13,19,29)$.