Ta có $d=\frac{1}{2}d\left( H;\left( SCD \right) \right)$

Kẻ $\left\{ \begin{align}  & HE\bot CD \\  & HP\bot SE \\ \end{align} \right.\Rightarrow d\left( H;\left( SCD \right) \right)=HP\Rightarrow d=\frac{HP}{2}$

Góc $SCH={{60}^{0}}\Rightarrow \tan {{60}^{0}}=\frac{SH}{HC}=\sqrt{3}\Rightarrow SH=HC\sqrt{3}$

Cạnh $HC=\sqrt{B{{C}^{2}}+B{{H}^{2}}}=\sqrt{{{a}^{2}}+\le {{\left( \frac{2a}{3} \right)}^{2}}}=\frac{a\sqrt{13}}{3}$

$\Rightarrow SH=a\sqrt{\frac{13}{3}}\Rightarrow \frac{1}{H{{P}^{2}}}=\frac{1}{S{{H}^{2}}}+\frac{1}{H{{E}^{2}}}=\frac{3}{13{{a}^{2}}}+\frac{1}{{{a}^{2}}}=\frac{16}{13{{a}^{2}}}$

$\Rightarrow HP=\frac{a\sqrt{13}}{4}\Rightarrow d=\frac{HP}{2}=\frac{a\sqrt{13}}{8}$