Gọi $O$ là tâm của $ΔABC$
$\to OA = OB =OC = \dfrac{AB\sqrt3}{3} = \dfrac{a\sqrt3}{3}$
Ta có:
$SO\perp (ABC)$ (hình chóp đều)
$\to SO\perp OA$
$\to SO = \sqrt{SA^2 - OA^2} = \sqrt{4a^2 - \dfrac{a^2}{3}}=\dfrac{a\sqrt{33}}{3}$
$\to V_{S.ABC} = \dfrac{1}{3}S_{ABC}.SO = \dfrac{1}{3}\cdot\dfrac{a^2\sqrt3}{4}\cdot \dfrac{a\sqrt{33}}{3} = \dfrac{a^3\sqrt{11}}{12}$
Ta được:
$\dfrac{V_{S.AMN}}{V_{S.ABC}} = \dfrac{SA}{SA}\cdot\dfrac{SM}{SB}\cdot\dfrac{SN}{SC} = 1\cdot\dfrac{1}{2}\cdot\dfrac{3}{4} = \dfrac{3}{8}$
$\to \dfrac{V_{A.BCNM}}{V_{S.ABC}} =\dfrac{5}{8}$
$\to V_{A.BCNM} = \dfrac{5}{8}V_{S.ABC} = \dfrac{5}{8}\cdot\dfrac{a^3\sqrt{11}}{12} = \dfrac{5a^3\sqrt{11}}{96}$
