(15) (50 points) Let integers $a, b, c$ satisfy $1 \leqslant a \leqslant b \leqslant c$, and $a \mid b+c+1$, $b \mid c+a+1$, $c \mid a+b+1$. Find all triples $(a, b, c)$.
(15) Case $1 \quad a \leqslant b<c$, at this time
$$
a+b+1 \leqslant a+c<2 c .
$$

Also, $c \mid a+b+1$, so $a+b+1=c$. Substituting $c=a+b+1$ into $a \mid b+c+1$ and $b \mid c+a+1$, we get $a \mid a+2b+2, b \mid 2a+b+2$. That is, $a \mid 2b+2, b \mid 2a+2$.
From $2a+2 \leqslant 4a \leqslant 4b$ we know $2a+2=b, 2b, 3b, 4b$.
From $2a+2=b$ and $a \mid 2b+2$ we know $a \mid 4a+6$, i.e., $a \mid 6$.
At this time $(a, b, c)=(1,4,6),(2,6,9),(3,8,12),(6,14,21)$.
From $2a+2=2b$ and $a \mid 2b+2$ we know $a \mid 2a+4$, i.e., $a \mid 4$.
At this time $(a, b, c)=(1,2,4),(2,3,6),(4,5,10)$.
From $2a+2=3b$ and $a \mid 2b+2$ we know $3 \mid a+1, 3a \mid 4a+10$, so $a \mid 10, 3 \mid a+1$.
Therefore, $(a, b, c)=(2,2,5)$.
From $2a+2=4b$ and $a \leqslant b$ we know $a=b=1$, at this time $(a, b, c)=(1,1,3)$.
Case $2a \leqslant b=c$, at this time from $b \mid c+a+1$, we know $b \mid a+1$.
Also, $1 \leqslant a \leqslant b$, so $a+1=b, 2b$.
If $a+1=b$, then from $a \mid b+c+1$ we know $a \mid 2a+3$, i.e., $a \mid 3$.
At this time $(a, b, c)=(1,2,2),(3,4,4)$.
If $a+1=2b$, then from $1 \leqslant a \leqslant b$ we know $a=b=1$.
At this time $(a, b, c)=(1,1,1)$.
In summary, $(a, b, c)=(1,1,1),(1,2,2),(3,4,4),(1,1,3),(2,2,5),(1,2,4),(2,3,6),(4,5,10),(1,4,6)$
$$
\begin{array}{l}
(2,6,9),(3,8,12),(6,14,21)
\end{array}
$$