We have that \[\cos^2 A = 1 - \sin^2 A = \frac{16}{25},\]so $ \cos A = \pm \frac{4}{5} $. Similarly, \[\cos^2 B = 1 - \sin^2 B = \frac{49}{625},\]so $ \cos B = \pm \frac{7}{25} $. Then \begin{align*} \sin C &= \sin (180^\circ - A - B) \\ &= \sin (A + B) \\ &= \sin A \cos B + \cos A \sin B \\ &= \frac{3}{5} \left( \pm \frac{7}{25} \right) + \left( \pm \frac{4}{5} \right) \frac{24}{25}.\end{align*}The possible values of this expression are $ \pm \frac{3}{5} $ and $ \pm \frac{117}{125} $. But $ \sin C $ must be positive, so the possible values of $ \sin C $ are $ \boxed{\frac{3}{5}, \frac{117}{125}} $.