6. Find all triples $(p, q, r)$ satisfying the following conditions:
(1) $p, q, r \in \mathbf{N}^{\cdot}, p \geqslant q \geqslant r$;
(2) At least two of $p, q, r$ are prime numbers;
(3) $\frac{(p+q+r)^{2}}{p q r}$ is a positive integer.
6. Let $(a, b, c)$ be a solution that meets the problem's conditions, and $a, b$ are prime numbers, $a \geqslant b$. Since $\frac{(a+b+c)^{2}}{a b c} \in \mathbf{N}^{*}$, and $a, b$ are prime numbers, we have $a|b+c, b| c+a, c \mid(a+b)^{2}$.
(1) Suppose $a=b$. Then from $a \mid b+c$, we get $a \mid c$, and from $c \mid(a+b)^{2}$, we get $c \mid 4 a^{2}$.
Thus, $c$ can be $a, 2 a, 4 a, a^{2}, 2 a^{2}, 4 a^{2}$.
Substituting these into $\frac{(a+b+c)^{2}}{a b c}$, we find: when $c=a$, $a=3$; when $c=2 a$, $a=2$; when $c=4 a$, $a=3$; 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) Suppose $a>b$. From $a \mid b+c$, we can set $b+c=k a\left(k \in \mathbf{N}^{*}\right)$. Then $c=k a-b, c+a=(k+1) a-b$,
Since $b \mid c+a$, we have $b \mid(k+1) a$,
But $a, b$ are different prime numbers, so $b \mid k+1$, we can set $k+1=m b\left(m \in \mathbf{N}^{*}\right), c=(m b-1) a-b=m a b-$ $a-b$,
Thus, $a+b+c=m a b$.
$\frac{(a+b+c)^{2}}{a b c}=\frac{m a b(a+b+c)}{a b c}=m+\frac{m(a+b)}{c} \in \mathbf{N}^{*}$.
Therefore, set $m(a+b)=n c(c \in \mathbf{N})$, then $m(a+b)=n c=m n a b-n a-n b$,
That is, $m n a b=(m+n)(a+b)$.
Since $a, b$ are different prime numbers, we must have $a|m+n, b| m+n$, hence $a b \mid m+n$.
Let $m+n=t a b\left(t \in \mathbf{N}^{*}\right)$, we have $m n=t(a+b)$.
Since $t(a+b)-t a b+1=m n-(m+n)+1=(m-1)(n-1) \geqslant 0 \quad$ (because $m, n \in \mathbf{N}^{*}$)
We get $t(a b-a-b) \leqslant 1$, i.e., $t[(a-1)(b-1)-1] \leqslant 1$.
Since $t, a, b \in \mathbf{N}^{*}$, and $a>b$, we must have $(a-1)(b-1)-1=1$.
Thus, $a=3, b=2$, at this time $t=1, m n=5, m+n=6$,
Solving gives $m=1$ or $5, c=m a b-a-b=1$ or 25.
Therefore, $(a, b, c)=(3,2,1),(3,2,25)$.
In summary, there are 5 solutions that meet the problem's conditions, which are $(3,3,3),(2,2,4),(3,3,12),(3,2,1),(3,2,25)$.