Cho hình chóp S.ABC có đáy ABC là tam giác vuông cân tại A. Tam giác

Gọi H là trung điểm của AB do tam giác SAB đều nên $SH\perp AB.$

Ta có: $\left\{\begin{array}{l}\left(SAB\right)\perp \left(ABC\right)=AB\\ SH\subset \left(SAB\right),SH\perp AB\end{array}\right.\Rightarrow SH\perp \left(ABC\right).$

Dựng hình vuông ABFC ta có $BF//AC\Rightarrow AC//\left(BMF\right)$

$\Rightarrow d\left(AC;BM\right)=d\left(AC;\left(BMF\right)\right)=d\left(A;\left(BMF\right)\right)$.

Lại có $AH\cap \left(BMF\right)=B\Rightarrow \frac{d\left(A;\left(BMF\right)\right)}{d\left(H;\left(BMF\right)\right)}=\frac{AB}{HB}=2\Rightarrow d\left(A;\left(BMF\right)\right)=2d\left(H;\left(BMF\right)\right)$

Trong (SHC) kẻ $ME//SH\left(E\in CH\right)\Rightarrow ME\perp \left(ABC\right).$

Kéo dài HC cắt BF tại N, áp dụng định lí Ta-lét ta có $\frac{BH}{FC}=\frac{NH}{NC}=\frac{NB}{NF}=\frac{1}{2}\Rightarrow H$ là trung điểm của NC.

$\Rightarrow ACBN$ là hình bình hành

Ta có: $HE\cap \left(BMF\right)=N\Rightarrow \frac{d\left(H;\left(BMF\right)\right)}{d\left(E;\left(BMF\right)\right)}=\frac{HN}{EN}=\frac{HC}{HE+HC}$

Áp dụng định lí Ta-lét ta có: $\frac{HE}{HC}=\frac{SM}{SC}=\frac{1}{3}\Rightarrow HE=\frac{1}{3}HC$

$\Rightarrow \frac{d\left(H;\left(BMF\right)\right)}{d\left(E;\left(BMF\right)\right)}=\frac{HN}{EN}=\frac{HC}{HE+HC}=\frac{HC}{\frac{1}{3}HC+HC}=\frac{3}{4}\Rightarrow d\left(H;\left(BMF\right)\right)=\frac{3}{4}d\left(E;\left(BMF\right)\right)$

$\Rightarrow d\left(AC;BM\right)=d\left(A;\left(BMF\right)\right)=\frac{3}{2}d\left(E;\left(BMF\right)\right)$

Trong (ABC) kẻ $EI//AB\left(I\in BF\right),$ trong (MEI) kẻ $EJ\perp IM$ ta có:

$\left\{\begin{array}{l}BF\perp AB,EI//AB\Rightarrow BF\perp EI\\ BF\perp ME\left(ME\perp \left(ABC\right)\right)\end{array}\right.\Rightarrow BF\perp \left(MEI\right)\Rightarrow BF\perp EJ$

$\left\{\begin{array}{l}EJ\perp BF\\ EJ\perp MI\end{array}\right.\Rightarrow EJ\perp \left(BMF\right)\Rightarrow d\left(E;\left(BMF\right)\right)=EJ$

$\Rightarrow d\left(AC;BM\right)=\frac{3}{2}d\left(E;\left(BMF\right)\right)=\frac{3}{2}EJ=\frac{4\sqrt{21}}{7}\Rightarrow EJ=\frac{8}{\sqrt{21}}.$

Ta có: $\frac{ME}{SH}=\frac{CM}{CS}=\frac{2}{3}\Rightarrow ME=\frac{2}{3}SH.$ Mà $SH=\frac{AB\sqrt{3}}{2}\Rightarrow ME=\frac{AB\sqrt{3}}{3}.$

Ta có: $\frac{HN}{EN}=\frac{3}{4}\left(cmt\right)\Rightarrow \frac{NE}{NH}=\frac{4}{3}\Rightarrow \frac{NE}{NC}=\frac{2}{3}=\frac{IE}{FC}=\frac{IE}{AB}\Rightarrow IE=\frac{2}{3}AB.$

Áp dụng hệ thức lượng trong tam giác MEI ta có:

$\frac{1}{E{J}^2}=\frac{1}{E{I}^2}+\frac{1}{E{M}^2}$

$\Rightarrow \frac{21}{64}=\frac{1}{\frac{4}{9}A{B}^2}+\frac{1}{\frac{1}{3}A{B}^2}$

$\Rightarrow \frac{21}{64}=\frac{21}{4A{B}^2}$

$\iff AB=4$

$\Rightarrow ME=\frac{AB\sqrt{3}}{3}=\frac{4\sqrt{3}}{3},S_{\Delta ABC}=\frac{1}{2}A{B}^2=8.$

Vậy $V_{C.ABM}=\frac{1}{3}ME.S_{\Delta ABC}=\frac{1}{3}.\frac{4\sqrt{3}}{3}.8=\frac{32\sqrt{3}}{9}.$

Chọn B.