Vì tam giác ABC vuông cân tại C, \(AB=3a\Rightarrow CA=CB=\frac{3a}{\sqrt{2}}\)
Gọi M là trung điểm \(AC\Rightarrow MC=\frac{3a}{2\sqrt{2}}\Rightarrow MB=\frac{3a\sqrt{5}}{2\sqrt{2}}\)
\(\Rightarrow BG=\frac{2}{3}BM=\frac{a\sqrt{5}}{\sqrt{2}}\Rightarrow SG=\sqrt{SB^2-BG^2}=a\)
\(\Rightarrow V_{S.ABC}=\frac{1}{3}SG.S_{\Delta ABC}=\frac{3a^2}{4}=\frac{3a^2}{4}\)
Kẻ \(GI\perp AC\left(I\in AC\right)\Rightarrow AC\perp\left(SGI\right)\)
Ta có : \(GI=\frac{1}{3}BC=\frac{a}{\sqrt{2}}\)
Kẻ \(GH\perp SI\left(H\in SI\right)\Rightarrow GH\perp\left(SAC\right)\Rightarrow d\left(G,\left(SAC\right)\right)=GH\)
Ta có \(\frac{1}{GH^2}=\frac{1}{GS^2}+\frac{1}{GI^2}\Rightarrow GH=\frac{a}{\sqrt{3}}\Rightarrow3d\left(B,\left(SAC\right)\right)=3GH=a\sqrt{3}\)
