Ta có: \(\frac{{{V_{ABCD}}}}{{{V_{AB'C'D'}}}} = \frac{{AB}}{{AB'}}.\frac{{AC}}{{AC'}}.\frac{{AD}}{{AD'}} \le {\left( {\frac{{\frac{{AB}}{{AB'}} + \frac{{AC}}{{AC'}} + \frac{{AD}}{{AD'}}}}{3}} \right)^3} = \frac{{512}}{{27}}.\)

\( \Rightarrow {V_{AB'C'D'}} \ge \frac{{27}}{{512}}{V_{ABCD}}\).

Dấu bằng xảy ra khi \(\frac{{AB}}{{AB'}} = \frac{{AC}}{{AC'}} = \frac{{AD}}{{AD'}} = \frac{8}{3} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{\left( {B'C'D'} \right)//(BCD)}\\{\overrightarrow {AB'} = \frac{3}{8}\overrightarrow {AB} }\end{array}} \right.\) .

Ta có: \(\left\{ {\begin{array}{*{20}{l}}{\overrightarrow {BC} = ( - 1; - 3;3)}\\{\overrightarrow {BD} = (2;4; - 6)}\end{array} \Rightarrow {{\vec n}_{\left( {B'C'D'} \right)}} = [\overrightarrow {BC} ,\overrightarrow {BD} ] = (6;0;2)} \right.\).

Tiếp tục có: \(\overrightarrow {AB} = (1;1;1)\) nên \(\overrightarrow {AB'} = \frac{3}{8}\overrightarrow {AB} = \left( {\frac{3}{8};\frac{3}{8};\frac{3}{8}} \right) \Rightarrow B'\left( {\frac{{11}}{8};\frac{3}{8}; - \frac{{13}}{8}} \right)\).

Khi đó phương trình mặt phẳng \(\left( {B'C'D'} \right)\) là

\(6\left( {x - \frac{{11}}{8}} \right) + 2\left( {z + \frac{{13}}{8}} \right) = 0 \Leftrightarrow 6x + 2z - 5 = 0\).

Vậy \(m = 0,n = 2,p = - 5 \Rightarrow {m^2} - n - p = 3\). Chọn A.