Đáp án: $24$

Giải thích các bước giải:

Kẻ $GE\perp AC$

Ta có:

$A'G\perp (ABC)$

$\to A'G\perp AC$

$\to AC\perp (GEA')$

$\to \widehat{A'EG}=60^o$

Vì $G$ là trọng tâm $\Delta ABC$

$S_{GAC}=\dfrac13S_{ABC}$

$\to \dfrac12GE\cdot AC=\dfrac13\cdot \dfrac12\cdot BA\cdot BC$

$\to \dfrac12\cdot GE\cdot 5=\dfrac13\cdot \dfrac12\cdot 3\cdot \sqrt{5^2-3^2}$

$\to GE=\dfrac45$

Ta có:

$[A; C'A'; B']=60^o$

$\to \widehat{(A'C'CA, ABC)}=60^o$

$\to \widehat{A'GE}=60^o$

$\to A'G=GE\sqrt3=\dfrac{4\sqrt3}5$

$\to V_{ABC.A'B'C'}=\dfrac{4\sqrt3}5\cdot \dfrac12\cdot 3\cdot 4=\dfrac{24\sqrt{3}}{5}$

$\to x=24$