Phương pháp giải:

+) Kẻ \(CH \bot AB\); \(CK \bot SB\), chứng minh \(\angle \left( {\left( {SAB} \right);\left( {SBC} \right)} \right) = \angle \left( {HK;CK} \right) = \angle CKH = {60^0}\).

+) Chứng minh \(\Delta BHK \sim \Delta BSA\,\,\left( {g - g} \right) \Rightarrow \frac{{HK}}{{SA}} = \frac{{HB}}{{SB}}\), từ đó tính HK.

Lời giải chi tiết:

Trong \(\left( {ABC} \right)\) kẻ \(CH \bot AB\) ta có: \(\left\{ \begin{array}{l}CH \bot AB\\CH \bot SA\end{array} \right. \Rightarrow CH \bot \left( {SAB} \right) \Rightarrow CH \bot SB\)

Trong (SBC) kẻ \(CK \bot SB\) ta có: \(\left\{ \begin{array}{l}CH \bot SB\\CK \bot SB\end{array} \right. \Rightarrow SB \bot \left( {CHK} \right) \Rightarrow HK \bot SB\).

\( \Rightarrow \angle \left( {\left( {SAB} \right);\left( {SBC} \right)} \right) = \angle \left( {HK;CK} \right) = \angle CKH = {60^0}\).

Xét tam giác vuông ABC ta có: \(BC = \sqrt {4{a^2} - {a^2}}  = a\sqrt 3 \)

\(CH = \frac{{AC.BC}}{{AB}} = \frac{{a\sqrt 3 .a}}{{2a}} = \frac{{a\sqrt 3 }}{2}\).

Xét tam giác vuông \(CHK\) có : \(HK = HC.\cot {60^0} = \frac{{a\sqrt 3 }}{2}.\frac{1}{{\sqrt 3 }} = \frac{a}{2}\).

\(HB = \frac{{B{C^2}}}{{AB}} = \frac{{3{a^2}}}{{2a}} = \frac{{3a}}{2}\)

Ta có \(\Delta BHK \sim \Delta BSA\,\,\left( {g.g} \right) \Rightarrow \frac{{HK}}{{SA}} = \frac{{HB}}{{SB}}\)

\(\begin{array}{l} \Rightarrow \frac{{\frac{a}{2}}}{{SA}} = \frac{{\frac{{3a}}{2}}}{{\sqrt {S{A^2} + 4{a^2}} }} \Rightarrow 3SA = \sqrt {S{A^2} + 4{a^2}} \\ \Leftrightarrow 9S{A^2} = S{A^2} + 4{a^2} \Leftrightarrow 8S{A^2} = 4{a^2} \Leftrightarrow SA = \frac{{a\sqrt 2 }}{2}\end{array}\) 

Vậy \({V_{S.ABC}} = \frac{1}{3}SA.{S_{\Delta ABC}} = \frac{1}{3}.\frac{{a\sqrt 2 }}{2}.\frac{1}{2}.a.a\sqrt{3} = \frac{{{a^3}\sqrt 6 }}{12}\).

Chọn D.