Vì $S.ABC$ là chóp đều nên: 

$SG \bot \left( {ABC} \right) \Rightarrow SG \bot AB \subset \left( {ABC} \right)$

Tam giác $BAC$ đều nên $AM\bot MC$ do $M$ là trung điểm của $AB$
$\begin{array}{l}
\left\{ \begin{array}{l}
AB \bot MC\\
AB \bot SG\\
SG \cap MC = \left\{ G \right\}
\end{array} \right. \Rightarrow AB \bot \left( {SMC} \right)\\
b)GC = \dfrac{2}{3}GM = \dfrac{2}{3}.\dfrac{{a\sqrt 3 }}{2} = \dfrac{{a\sqrt 3 }}{3}\\
 \Rightarrow SG = \sqrt {S{C^2} - G{C^2}}  = \dfrac{{a\sqrt {33} }}{3}\\
 \Rightarrow {V_{S.ABC}} = \dfrac{1}{3}SG.{S_{ABC}} = \dfrac{1}{3}.\dfrac{{a\sqrt {33} }}{3}\dfrac{{{a^2}\sqrt 3 }}{4}\\
 = \dfrac{{{a^3}\sqrt {11} }}{{12}}\\
c)\dfrac{{{d_{\left( {A,\left( {SBC} \right)} \right)}}}}{{{d_{\left( {G,\left( {SBC} \right)} \right)}}}} = \dfrac{{GK}}{{AK}} = 3\left( {K = AG \cap BC} \right)\\
 \Rightarrow {d_{\left( {A,\left( {SBC} \right)} \right)}} = 3{d_{\left( {G,\left( {SBC} \right)} \right)}}\\
 \bullet GH \bot SK\left( {H \in SK} \right)\\
\left\{ \begin{array}{l}
BC \bot SG\\
BC \bot GK
\end{array} \right. \Rightarrow BC \bot \left( {SGK} \right) \Rightarrow BC \bot GH\\
GH \bot SK \Rightarrow GH \bot \left( {SBC} \right) \Rightarrow {d_{\left( {G,\left( {SBC} \right)} \right)}} = GH\\
GH = \dfrac{{SG.GK}}{{\sqrt {S{G^2} + G{K^2}} }} = \dfrac{{\dfrac{{a\sqrt {33} }}{3}.\dfrac{{a\sqrt 3 }}{6}}}{{a\sqrt {\dfrac{{33}}{9} + \dfrac{3}{{36}}} }} = \dfrac{{a\sqrt {165} }}{{45}}
\end{array}$