Kẻ \[{\rm{BH}} \bot {\rm{AC}}\,\,\left( {{\rm{H}} \in {\rm{AC}}} \right);\,\,\,{\rm{HK}} \bot {\rm{SA}}\,\,\left( {{\rm{K}} \in {\rm{SA}}} \right)\].

Khi đó \(\left\{ {\begin{array}{*{20}{l}}{{\rm{SA}} = ({\rm{SAC}}) \cap ({\rm{SAB}})}\\{{\rm{HK}} \bot {\rm{SA}},{\rm{HK}} \subset ({\rm{SAC}})}\\{{\rm{BK}} \bot {\rm{SA}},{\rm{BK}} \subset ({\rm{SAB}})}\end{array}} \right.\)

\( \Rightarrow (({\rm{SAC}}),({\rm{SAB}})) = ({\rm{HK}},{\rm{BK}}) = \widehat {{\rm{HKB}}} = \alpha \)

Ta có: \(BC = \sqrt {A{C^2} - A{B^2}}  = a\sqrt 2 \).

Khi đó \(\cos \alpha  = \sqrt {\frac{6}} \,;\,\,\alpha  \in \left[ {0\,;\,\,90^\circ } \right] \Rightarrow \sin \alpha  = \sqrt {\frac} \).

Do đó \(\sin \alpha  = \frac{{{\rm{BH}}}}{{{\rm{BK}}}} \Rightarrow {\rm{BK}} = \frac{{{\rm{BH}}}}{{\sin \alpha }} = \frac{{\frac{{{\rm{a}}\sqrt 2 }}{{\sqrt 3 }}}}{{\sqrt {\frac} }} = \frac{{{\rm{a}}\sqrt {38} }}{{\sqrt {39} }}\).

Ta có: \(\left\{ {\begin{array}{*{20}{l}}{AB \bot SC}\\{AB \bot BC}\end{array} \Rightarrow AB \bot \left( {SBC} \right) \Rightarrow AB \bot SB} \right.\).

Khi đó tam giác \[SAB\] vuông tại \[B.\] Do đó \(\frac{1}{{{\rm{B}}{{\rm{K}}^2}}} = \frac{1}{{{\rm{A}}{{\rm{B}}^2}}} + \frac{1}{{{\rm{S}}{{\rm{B}}^2}}} \Rightarrow {\rm{S}}{{\rm{B}}^2} = 38{{\rm{a}}^2} \Rightarrow {\rm{SB}} = \sqrt {38} {\rm{a}}\).

Tam giác \({\rm{SBC}}\) vuông tại \({\rm{C}}\). Khi đó \({\rm{SC}} = \sqrt {{\rm{S}}{{\rm{B}}^2} - {\rm{B}}{{\rm{C}}^2}}  = 6a\). Đáp án: \[{\bf{6a}}\].