Đáp án:

$\dfrac{7\sqrt6}{4}$

Giải thích các bước giải:

$V_{A.BCD}=\dfrac{\left(2\sqrt3\right)^3\sqrt2}{12}= 2\sqrt6$

$d(A;(BCD))= \dfrac{\left(2\sqrt3\right)\sqrt6}{3} = 2\sqrt2$

Ta có:

$\bullet\quad \overrightarrow{MA} + \overrightarrow{MB}=\overrightarrow{0}$

$\Rightarrow MA = MB =\dfrac12AB$

$\Rightarrow d(M;(BCD))=\dfrac12d(A;(BCD))= \sqrt2$

$\bullet\quad \overrightarrow{NB} + 3\overrightarrow{NC}=\overrightarrow{0}$

$\Rightarrow CN =\dfrac34BC = \dfrac{3\sqrt3}{2}$

$\bullet\quad \overrightarrow{PC} + 2\overrightarrow{PD}=\overrightarrow{0}$

$\Rightarrow CP =\dfrac13CD = \dfrac{2\sqrt3}{3}$

Ta được:

$\bullet\quad  S_{CNP}=\dfrac12CN.CP.\sin\widehat{NCP}$

$\Leftrightarrow S_{CNP}=\dfrac12\cdot \dfrac{3\sqrt3}{2}\cdot \dfrac{2\sqrt3}{3}\cdot \sin60^\circ$

$\Leftrightarrow S_{CNP}= \dfrac{3\sqrt3}{4}$

$\bullet\quad V_{M.CNP}=\dfrac13S_{CNP}.d(M;(CNP))$

$\Leftrightarrow V_{M.CNP}= \dfrac13\cdot \dfrac{3\sqrt3}{4}\cdot \sqrt2$

$\Leftrightarrow V_{M.CNP}=\dfrac{\sqrt6}{4}$

Khi đó:

$V_{A.BNPD}= V_{A.BCD}- V_{M.CNP}= \dfrac{7\sqrt6}{4}$