\(V_{SABC}=\dfrac{1}{6}SA.AB.BC=\dfrac{a^3}{6}\)
b.
\(\left\{{}\begin{matrix}SA\perp\left(ABC\right)\\BC\perp AB\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\Rightarrow d\left(C;\left(SAB\right)\right)=BC=a\)
M là trung điểm SA \(\Rightarrow S_{SBM}=\dfrac{1}{2}S_{SAB}=\dfrac{1}{4}.SA.AB=\dfrac{a^2}{4}\)
\(SN=2NC\Rightarrow SN=\dfrac{2}{3}SC\)
\(\Rightarrow d\left(N;\left(SAB\right)\right)=\dfrac{2}{3}d\left(C;\left(SAB\right)\right)=\dfrac{2a}{3}\)
\(\Rightarrow V_{SMBN}=\dfrac{1}{3}.\dfrac{2a}{3}.\dfrac{a^2}{4}=\dfrac{a^3}{18}\)
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