Chọn: Đáp án A

Từ $C$ dựng$CI\parallel DE\Rightarrow CE=DI=\frac{a}{2}$ và$DE\parallel \left( SCI \right)$

$\Rightarrow d\left( DE,SC \right)=d\left( DE,\left( CIS \right) \right)$

Từ$A$ kẻ$AK\bot CI$ cắt $ED$ tại$H$ , cắt$CI$ tại $K$

Ta có:$\left\{ \begin{align}  & SA\bot CI \\  & AK\bot CI \\ \end{align} \right.\Rightarrow CI\bot \left( SCI \right)\bot \left( SAK \right)$ theo giao tuyến$SK$

Trong mặt phẳng$\left( SAL \right)$ kẻ$HT\bot AK\Rightarrow HT\bot \left( SCI \right)\Rightarrow d\left( DE,SC \right)=d\left( H,\left( SCI \right) \right)=HT$

Ta có:${{S}_{ACI}}=\frac{1}{2}AK.CI=\frac{1}{2}CD.AI\Rightarrow AK=\frac{CD.AI}{CI}=\frac{a.\frac{3}{2}a}{\sqrt{{{a}^{2}}+{{\left( \frac{a}{2} \right)}^{2}}}}=\frac{3a}{\sqrt{5}}$

Kẻ$KM\parallel AD\left( M\in ED \right)\Rightarrow \frac{HK}{HA}=\frac{KM}{AD}=\frac{1}{2}\Rightarrow HK=\frac{1}{3}AK=\frac{a}{\sqrt{5}}$

Lại có:$\sin \widehat{SKA}=\frac{SA}{SK}=\frac{HT}{HK}\Rightarrow HT=\frac{SA.HK}{SK}=\frac{a\sqrt{2}.\frac{a}{\sqrt{5}}}{\sqrt{2{{a}^{2}}+\frac{9{{a}^{2}}}{5}}}=\frac{\sqrt{38}}{19}a$

Vậy$d\left( ED,SC \right)=\frac{\sqrt{38}}{19}a$