Chọn A
Giả sử \(d\left(M,\left(OBC\right)\right)=a, d\left(M,\left(OAC\right)\right)=b, d\left(M,\left(OAB\right)\right)=c.\)
Suy ra thể tích khối hộp bằng V=abc
Mặt khác \(V_{OABC} =V_{M.OAB} +V_{M.OAC} +V_{M.OBC} \Leftrightarrow \frac{1}{6} .3.6.12=\frac{1}{6} .3.6.c+\frac{1}{6} 3.12.b+\frac{1}{6} .6.12.a\)
\(\[\Leftrightarrow \frac{a}{3} +\frac{b}{6} +\frac{c}{12} =1 \Rightarrow 1\ge 3\sqrt[{3}]{\frac{a}{3} \frac{b}{6} \frac{c}{12} } \Rightarrow a.b.c\le \frac{12.6.3}{27} =8\left({\rm cm}^{{\rm 3}} \right).\] \)
Dấu bằng xảy ra khi\( \frac{a}{3} =\frac{b}{6} =\frac{c}{12} =\frac{1}{3} \Leftrightarrow a=1,\, \, b=2,\, \, c=4.\)