Solubility of $\mathrm{M}{\left(\mathrm{OH}\right)}_3$ in 0.1 M $\mathrm{NaOH}$ is  $\left({\mathrm{K}}_{\mathrm{sp}}\mathrm{of} \mathrm{M}{\left(\mathrm{OH}\right)}_3=9\times {10}^{-24}{\mathrm{M}}^4\right)$

$$\begin{gathered} \mathrm{M}{\left(\mathrm{OH}\right)}_3\leftrightharpoons \underset{\mathrm{S}}{{\mathrm{M}}^{+3}}+\underset{3\mathrm{S}}{3{\mathrm{OH}}^{-}} \\ \mathrm{NaOH}\to \underset{0.1 \mathrm{M}}{{\mathrm{Na}}^{+}}+\underset{0.1 \mathrm{M}}{{\mathrm{OH}}^{-}} \\ {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{M}}^{+3}\right]{\left[{\mathrm{OH}}^{-}\right]}^3 \\ {\mathrm{K}}_{\mathrm{sp}}=\mathrm{S}{\left(3\mathrm{S}+0.1\right)}^3; \text{'}\mathrm{S}\text{'} \mathrm{too} \mathrm{small} \mathrm{compared} \mathrm{to} 0.1\mathrm{M}, \mathrm{it} \mathrm{can} \mathrm{be} \mathrm{neglected}; \\ \mathrm{S}=\frac{{\mathrm{K}}_{\mathrm{sp}}}{{\left({10}^{-1}\right)}^3}=\frac{9\times {10}^{-24}}{{10}^{-3}}=9\times {10}^{-21} \mathrm{M} \end{gathered}$$