II. The side lengths $a, b, c (a \leqslant b \leqslant c)$ of $\triangle A B C$ simultaneously satisfy the following three conditions:
(1) $a, b, c$ are all integers;
(2) $a, b, c$ form a geometric sequence;
(3) At least one of $a$ and $c$ equals 100.
Find all possible solutions for the triplet $(a, b, c)$.
(ii) If $a=100$, then $b^{2}=100 c, \therefore 10 \mid b$.
Also, $100+b>c=\frac{b^{2}}{100}, b^{2}-100 b-100^{2}<0, b^{2}+100 b-100^{2}>0$, thus
$50(\sqrt{5}-1)<b \leqslant 100$.
Noticing that $10 \mid b$, then $b$ can take $70,80,90,100$. The corresponding $c=49,64,81,100$.

In summary, the triplets $(a, b, c)$ have 10 possible solutions:
$$
\begin{array}{l}
\quad(49,70,100),(64,80,100),(81,90,100), \\
(100,100,100),(100,110,121),(100,120,144), \\
(100,130,169),(100,140,196),(100,150,225), \\
(100,160,256) .
\end{array}
$$