Giải thích các bước giải:

Gọi $E$ là trung điểm $AC$

$\to BE\perp AC$ vì $\Delta ABC$ đều cạnh $a$

$\to BE=\dfrac{a\sqrt3}2$

Mà $BB'\perp (ABC)\to BB'\perp BE, BB'\perp AC$

$\to AC\perp B'EB$

Kẻ $BF\perp B'E$

$\to AC\perp BF$

$\to BF\perp (B'AC)$

$\to d(B, B'AC)=BF$

Mà $\dfrac1{BF^2}=\dfrac1{BE^2}+\dfrac1{BB'^2}$

$\to \dfrac1{BF^2}=\dfrac1{(\dfrac{a\sqrt3}2)^2}+\dfrac1{(2a)^2}$

$\to BF=\dfrac{2\sqrt3a}{\sqrt{19}}$

$\to d(B, B'AC)=\dfrac{2\sqrt3a}{\sqrt{19}}$

Gọi $BM\cap AB'=D$

$\to \dfrac{DM}{DB}=\dfrac{B'M}{AB}=\dfrac12$

$\to d(M, B'AC)=\dfrac12d(B, B'AC)=\dfrac{\sqrt3a}{\sqrt{19}}$