Kẻ $OH\bot CD\left( H\in CD \right)$, kẻ $OK\bot SH\left( K\in SH \right)$. Ta chứng minh được rằng $OK\bot \left( SCD \right)$

Vì $\frac{MO}{MC}=\frac{3}{2}\Rightarrow {{d}_{\left( M,\left( SCD \right) \right)}}=\frac{3}{2}{{d}_{\left( O,\left( SCD \right) \right)}}=\frac{3}{2}OK$

Trong tam giác SOH ta có: $OK=\sqrt{\frac{O{{H}^{2}}.O{{S}^{2}}}{O{{H}^{2}}+O{{S}^{2}}}}=\frac{a\sqrt{6}}{6}$

Vậy ${{d}_{\left( M,\left( SCD \right) \right)}}=\frac{3}{2}OK=\frac{a\sqrt{6}}{4}$