Lời giải chi tiết:

+ Xét \(\Delta ABC\): Kẻ \(AH \bot CB\)

Mà \(CB \bot SA\,\,\,\left( {SA \bot \left( {ABC} \right)} \right)\)\( \Rightarrow CB \bot \left( {SAH} \right)\)

+ Xét \(\Delta SAH\): Kẻ \(AK \bot SH\). Mà \(AK \bot BC\,\,\left( {BC \bot \left( {SAH} \right)} \right)\)

\( \Rightarrow AK \bot (SBC) \Rightarrow d\left( {A;\left( {SBC} \right)} \right) = AK = \dfrac{{a\sqrt {21} }}{7}\)

+ Xét \(\Delta SAH:\dfrac{1}{{A{K^2}}} = \dfrac{1}{{S{A^2}}} + \dfrac{1}{{A{H^2}}}\) (Hệ thức lượng)

\( \Leftrightarrow \dfrac{7}{{3{a^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{A{H^2}}} \Leftrightarrow A{H^2} = \dfrac{{3{a^2}}}{4} \Rightarrow AH = \dfrac{{\sqrt 3 }}{2}a\)

+ Xét \(\Delta ABC\) có: \(\dfrac{1}{{A{B^2}}} + \dfrac{1}{{A{C^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{{\left( {a\sqrt 3 } \right)}^2}}} = \dfrac{4}{3}{a^2}\)

Lại có: \(\dfrac{1}{{A{H^2}}} = \dfrac{1}{{{{\left( {\dfrac{{\sqrt 3 }}{2}a} \right)}^2}}} = \dfrac{4}{3}{a^2} \Rightarrow \dfrac{1}{{A{H^2}}} = \dfrac{1}{{A{B^2}}} + \dfrac{1}{{A{C^2}}}\)

\( \Rightarrow \Delta ABC\) vuông tại \(A\) có \(AH\) là đường cao.

+) Xét \(\Delta ABC:\,A{B^2} + A{C^2} = B{C^2}\)(Định lí Pytago) \( \Rightarrow BC = \sqrt {A{B^2} + A{C^2}}  = 2a.\)

\(\begin{array}{l} \Rightarrow {r_{day\,ABC}} = \dfrac{{BC}}{2} = \dfrac{{2a}}{2} = a\\ \Rightarrow R = \sqrt {\dfrac{{{h^2}}}{4} + {r^2}}  = \sqrt {\dfrac{{{a^2}}}{4} + {a^2}}  = \dfrac{{\sqrt 5 }}{2}a\end{array}\)

Chọn C.