Phương pháp giải:

Lời giải chi tiết:

Trong (ABCD) kẻ $BH \bot AC\,\,\left( {H \in AC} \right)$, ta có:

$\left\{ \begin{array}{l}\left( {ABCD} \right) \bot \left( {ACC'A'} \right) = AC\\AH \subset \left( {ABCD} \right),\,\,AH \bot AC\end{array} \right.$ $\Rightarrow AH \bot \left( {ACC'A'} \right)$.

Trong (ACC’A’) kẻ $HK \bot AA'\,\,\left( {K \in AA'} \right)$ ta có: $\left\{ \begin{array}{l}HK \bot AA'\\BH \bot AA'\end{array} \right.$ $AA' \bot \left( {BHK} \right) \Rightarrow AA' \bot BK$.

$\left\{ \begin{array}{l}\left( {ACC'A'} \right) \cap \left( {AA'B'B} \right) = AA'\\HK \subset \left( {ACC'A'} \right);\,\,HK \bot AA'\\BK \subset \left( {ACC'A'} \right);\,\,BK \bot AA'\end{array} \right.$ $\Rightarrow \angle \left( {\left( {ACC'A'} \right);\left( {AA'B'B} \right)} \right) = \angle \left( {HK;BK} \right) = \angle BKH = \alpha$.

Ta có: $BH \bot \left( {ACC'A'} \right) \Rightarrow BH \bot HK$ $\Rightarrow \Delta BHK$ vuông tại H.

Ta có: $BH = \dfrac{{AB.BC}}{{\sqrt {A{B^2} + B{C^2}} }} = \dfrac{{\sqrt 6 .\sqrt 3 }}{{\sqrt {6 + 3} }} = \sqrt 2$.

$\Rightarrow HK = BH.\cot \alpha  = \sqrt 2 .\dfrac{4}{3} = \dfrac{{4\sqrt 2 }}{3}$.

Áp dụng hệ thức lượng trong tam giác vuông ABC ta có: $AH = \dfrac{{A{B^2}}}{{AC}} = \dfrac{6}{{\sqrt {6 + 3} }} = 2$.

Xét tam giác vuông AHK có: $\sin \widehat {HAK} = \dfrac{{HK}}{{AH}} = \dfrac{{4\sqrt 2 }}{3}:2 = \dfrac{{2\sqrt 2 }}{3}$ $\Rightarrow \cos \widehat {HAK} =  \pm \sqrt {1 - {{\sin }^2}\widehat {HAK}}  =  \pm \dfrac{1}{3} = \cos \widehat {A'AC}$

Áp dụng định lí Cosin trong tam giác AA’C có:

TH1: $\cos \widehat {A'AC} = \dfrac{1}{3}$

$\begin{array}{l}\cos \widehat {A'AC} = \dfrac{{AA{'^2} + A{C^2} - A'{C^2}}}{{2AA'.AC}}\\ \Rightarrow \dfrac{1}{3} = \dfrac{{AA{'^2} + 9 - 9}}{{2.AA'.3}}\\ \Rightarrow \dfrac{1}{3} = \dfrac{{AA'}}{6} \Leftrightarrow AA' = 2\end{array}$

TH2: $\cos \widehat {A'AC} =  - \dfrac{1}{3}$

$\begin{array}{l}\cos \widehat {A'AC} = \dfrac{{AA{'^2} + A{C^2} - A'{C^2}}}{{2AA'.AC}}\\ \Rightarrow  - \dfrac{1}{3} = \dfrac{{AA{'^2} + 9 - 9}}{{2.AA'.3}}\\ \Rightarrow  - \dfrac{1}{3} = \dfrac{{AA'}}{6}\,\,\left( {ktm} \right)\end{array}$

$\begin{array}{l} \Rightarrow {S_{\Delta AA'C}} = \dfrac{1}{2}AA'.AC.\sin \widehat {A'AC}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}.2.3.\dfrac{{2\sqrt 2 }}{3} = 2\sqrt 2 \\ \Rightarrow {S_{ACC'A'}} = 2{S_{\Delta AA'C}} = 4\sqrt 2 \\ \Rightarrow {V_{B.ACC'A'}} = \dfrac{1}{3}BH.{S_{ACC'A'}} = \dfrac{1}{3}.\sqrt 2 .4\sqrt 2  = \dfrac{8}{3}\\ \Rightarrow {V_{ABC.A'B'C'}} = \dfrac{3}{2}{V_{B.ACC'A'}} = 4\\ \Rightarrow {V_{ABCD.A'B'C'D'}} = 2{V_{ABC.A'B'C'}} = 8.\end{array}$ 

Chọn C.