Ta có:
\(\begin{array}{l}\left. \begin{array}{l}SA \bot \left( {ABC} \right) \Rightarrow SA \bot BC\\AI \bot BC\end{array} \right\} \Rightarrow BC \bot \left( {SAI} \right)\\\left. \begin{array}{l} \Rightarrow BC \bot AH\\AH \bot SI\end{array} \right\} \Rightarrow AH \bot \left( {SBC} \right)\end{array}\)
Vậy \(d\left( {A,\left( {SBC} \right)} \right) = AH\).