71. Find all triples $(p, q, r)$ that satisfy the following conditions:
1) $p, q, r \in \mathbf{N}, p \geqslant q \geqslant r$;
2) At least two of $p, q, r$ are prime numbers;
3) $\frac{(p+q+r)^{2}}{p q r}$ is a positive integer.
Let $(a, b, c)$ be a solution that meets the problem's conditions, where $a$ and $b$ are prime numbers, and $a \geqslant b$. Using $\frac{(a+b+c)^{2}}{a b c} \in \mathbf{N}$ and the fact that $a$ and $b$ are prime, it is easy to deduce that $a|b+c, b| c+a, c \mid(a+b)^{2}$.
$1^{\circ}$ Suppose $a=b$.
In this case, from $a \mid b+c$ we get $a \mid c$. Then, from $c \mid(a+b)^{2}$, we get $c \mid 4 a^{2}$. Therefore, $c$ can be $a, 2a, 4a, a^2, 2a^2, 4a^2$. Substituting these into the equation, when $c=2a$, $a=2$; when $c=4a$, $a=3$; when $c=a^2$, $a=2$, all satisfy the problem's conditions. The corresponding solutions are:
$$
(a, b, c)=(3,3,3),(2,2,4),(3,3,12).
$$
$2^{\circ}$ Suppose $a>b$.
From $a \mid b+c$ we get $b+c=k a (k \in \mathbf{N})$, so $c=k a-b$, and $c+a=(k+1) a-b$. Then, from $b \mid a+c$ we know $b \mid (k+1) a$. Since $a$ and $b$ are different primes, we have $b \mid k+1$. Let $k+1=m b (m \in \mathbf{N})$, then $c=(m b-1) a-b=m a b-a-b$, and $a+b+c=m a b$. Therefore, $\frac{(a+b+c)^{2}}{a b c}=\frac{m a b(a+b+c)}{a b c}=m+\frac{m(a+b)}{c} \in \mathbf{N}$. Thus, $m(a+b)=n c=m n a b-n a-n b (n \in \mathbf{N})$, which simplifies to $m n a b=(m+n)(a+b)$. Since $a$ and $b$ are different primes, we must have $a \mid m+n$ and $b \mid m+n$, so $a b \mid m+n$. Let $m+n=t a b (t \in \mathbf{N})$, then $m n=t(a+b)$. Using $t(a+b)-t a b+1=m n-(m+n)+1=(m-1)(n-1) \geqslant 0$ (since $m, n \in \mathbf{N}$), we get $t(a b-a-b) \leqslant 1$, or $t[(a-1)(b-1)-1] \leqslant 1$. Since $t, a, b \in \mathbf{N}$ and $a > b$, we must have $(a-1)(b-1)-1=1$. Hence, $a=3, b=2$. At this point, $t=1, m n=5, m+n=6$, which gives $m=1$ or $5$, and $c=m a b-a-b=1$ or 25. Therefore, we get $(a, b, c)=(3,2,1),(3,2,25)$.

In summary, there are 5 different solutions to this problem: $(3,3,3)$, $(4,2,2)$, $(12,3,3)$, $(3,2,1)$, $(25,3,2)$.