Cho hình chóp S.ABC có đáy ABC là tam giác vuông với AB = AC = 2

Gọi N là trung điểm của BC ta có $MN//BC\Rightarrow BC//\left(AMN\right)\supset AM$

$\Rightarrow d\left(AM;BC\right)=d\left(BC;\left(AMN\right)\right)=d\left(C;\left(AMN\right)\right).$

Lại có $SC\cap \left(AMN\right)=M\Rightarrow \frac{d\left(C;\left(AMN\right)\right)}{d\left(S;\left(AMN\right)\right)}=\frac{CM}{SM}=1\Rightarrow d\left(C;\left(AMN\right)\right)=d\left(S;\left(AMN\right)\right)$

Ta có

$AM=\frac{1}{2}SC=\frac{1}{2}\sqrt{S{A}^2+A{C}^2}=\frac{1}{2}\sqrt{3^2+2^2}=\frac{\sqrt{13}}{2}$

$AN=\frac{1}{2}SB=\frac{1}{2}\sqrt{S{A}^2+A{B}^2}=\frac{1}{2}\sqrt{3^2+2^2}=\frac{\sqrt{13}}{2}$

$MN=\frac{1}{2}BC=\frac{1}{2}\sqrt{A{B}^2+A{C}^2}=\frac{1}{2}.2\sqrt{2}=\sqrt{2}$

Gọi p là nửa chu vi tam giác AMN ta có $p=\frac{\frac{\sqrt{13}}{2}+\frac{\sqrt{13}}{2}+\sqrt{2}}{2}=\frac{\sqrt{13}+\sqrt{2}}{2}.$

$\Rightarrow S_{\Delta AMN}=\sqrt{p\left(p-AM\right)\left(p-AN\right)\left(p-MN\right)}=\frac{\sqrt{22}}{4}.$

$\frac{V_{S.AMN}}{V_{S.ABC}}=\frac{SM}{SC}.\frac{SN}{SB}=\frac{1}{4}\Rightarrow V_{S.AMN}=\frac{1}{4}{V}_{S.ABC},V_{S.ABC}=\frac{1}{3}SA.\frac{1}{2}AB.AC=\frac{1}{6}\mathrm{.3.2.2}=2.$

$\Rightarrow V_{S.AMN}=\frac{1}{4}.2=\frac{1}{2}.$

Vậy $d\left(AM;BC\right)=d\left(S;\left(AMN\right)\right)=\frac{3V_{S.AMN}}{S_{\Delta AMN}}=\frac{3.\frac{1}{2}}{\frac{\sqrt{22}}{4}}=\frac{3\sqrt{22}}{11}.$

Chọn C.