21. A triangle whose angles are $A, B, C$ satisfies the following conditions
$$
\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\frac{12}{7},
$$
and
$$
\sin A \sin B \sin C=\frac{12}{25} .
$$

Given that $\sin C$ takes on three possible values $s_{1}, s_{2}$ and $s_{3}$, find the value of $100 s_{1} s_{2} s_{3}$ -
21. Answer. 48
By using factor formulae and double angle formulae:
$$
\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\frac{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}{1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}=\frac{12}{7},
$$
and
$$
\sin A \sin B \sin C=8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}=\frac{12}{25} .
$$

Solving these equations, we obtain
$$
\begin{array}{r}
\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=0.1 \\
\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}=0.6
\end{array}
$$

Furthermore,
$$
\sin \frac{C}{2}=\cos \left(\frac{A+B}{2}\right)=\cos \frac{A}{2} \cos \frac{B}{2}-\sin \frac{A}{2} \sin \frac{B}{2},
$$
multiplying both sides by $\sin \frac{C}{2} \cos \frac{C}{2}$, we get
$$
\sin ^{2} \frac{C}{2} \cos \frac{C}{2}=0.6 \sin \frac{C}{2}-0.1 \cos \frac{C}{2} .
$$
or equivalently,
$$
\left(1-t^{2}\right) t=0.6 \sqrt{1-t^{2}}-0.1 t \Leftrightarrow 11 t-10 t^{3}=6 \sqrt{1-t^{2}},
$$
where $t=\cos \frac{C}{2}$. This equation solves for $t=\sqrt{\frac{1}{2}}, \sqrt{\frac{4}{5}}, \sqrt{\frac{3}{10}}$, and so the corresponding values of $\sin C$ are
$$
1,0.8,0.6
$$
and hence $100 s_{1} s_{2} s_{3}=100 \cdot 1 \cdot 0.8 \cdot 0.6=48$.