Let $ 0 \le \alpha, \beta, \gamma \le \frac{\pi}{2}$ satisfy the conditions

$ \sin \alpha\plus{}\sin \beta\plus{}\sin \gamma\equal{}1, \sin \alpha \cos 2\alpha\plus{}\sin \beta\cos 2\beta\plus{}\sin \gamma \cos 2\gamma\equal{}\minus{}1.$

Find all possible values of $ \sin^2 \alpha\plus{}\sin^2 \beta\plus{}\sin^2 \gamma$.
1. Given the conditions:
   \[
   0 \le \alpha, \beta, \gamma \le \frac{\pi}{2}
   \]
   \[
   \sin \alpha + \sin \beta + \sin \gamma = 1
   \]
   \[
   \sin \alpha \cos 2\alpha + \sin \beta \cos 2\beta + \sin \gamma \cos 2\gamma = -1
   \]

2. For any \( x \in \left[0, \frac{\pi}{2} \right] \), we have:
   \[
   \cos 2x \ge -1 \quad \text{and} \quad \sin x \ge 0
   \]
   Therefore:
   \[
   \sin x \cos 2x \ge -\sin x
   \]
   with equality if and only if \( x = 0 \) or \( x = \frac{\pi}{2} \).

3. Summing up for \(\alpha, \beta, \gamma\):
   \[
   \sin \alpha \cos 2\alpha + \sin \beta \cos 2\beta + \sin \gamma \cos 2\gamma \ge -(\sin \alpha + \sin \beta + \sin \gamma) = -1
   \]
   Given that the equality holds, it implies:
   \[
   \sin \alpha \cos 2\alpha + \sin \beta \cos 2\beta + \sin \gamma \cos 2\gamma = -1
   \]
   This equality holds if and only if \(\alpha, \beta, \gamma \in \{0, \frac{\pi}{2}\}\).

4. Since \(\sin(0) = 0\) and \(\sin\left(\frac{\pi}{2}\right) = 1\), one of \(\alpha, \beta, \gamma\) must be \(\frac{\pi}{2}\) and the other two must be \(0\).

5. Therefore, the possible values of \(\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma\) are:
   \[
   \sin^2 \left(0\right) + \sin^2 \left(0\right) + \sin^2 \left(\frac{\pi}{2}\right) = 0 + 0 + 1 = 1
   \]

Conclusion:
\[
\boxed{1}
\]