Lời giải:

Ta có:

$H$ là trực tâm $∆ABC$ đều

$\Rightarrow \begin{cases}AH\perp BC\\BH\perp AC\end{cases}$

Lại có:

$SA\perp (ABC)\quad (gt)$

$\Rightarrow \begin{cases}MA\perp BC\\MA\perp BH\end{cases}$

Do đó:

$\begin{cases}BC\perp (MAH)\\BH\perp (MAC)\end{cases}$

$\Rightarrow \begin{cases}BC\perp HK\\BH\perp MC\end{cases}$

Mặt khác:

$K$ là trực tâm của $∆MBC$

$\Rightarrow BK\perp MC$

$\Rightarrow MC\perp (BHK)$

$\Rightarrow MC\perp HK$

$\Rightarrow HK\perp (MBC)$

b) Gọi $N$ là trung điểm $BC$

$\Rightarrow \begin{cases}AN\perp BC\\AN =\dfrac{a\sqrt3}{2}\end{cases}$

Ta có:

$HK\perp (MBC)$ (câu a)

$\Rightarrow HK\perp MN$

$\Rightarrow \widehat{H}=90^\circ$

$\Rightarrow ∆HKN\backsim ∆MAN\ (g.g)$

$\Rightarrow \dfrac{KN}{AN}=\dfrac{HN}{MN}$

$\Rightarrow KN =\dfrac{AN.HN}{MN}$

$\Rightarrow KN =\dfrac{\dfrac{a\sqrt3}{2}\cdot \dfrac{a\sqrt3}{6}}{\sqrt{\dfrac{3a^2}{4} + x^2}}$

$\Rightarrow KN =\dfrac{a^2}{4\sqrt{\dfrac{3a^2}{4} + x^2}}$

Trong mp$(MAN)$ kẻ $KI\perp AN$

$\Rightarrow KI//AM$

$\Rightarrow KI\perp (ABC)$

$\Rightarrow KI = d(K;(ABC))$

Áp dụng định lý $Thales$ ta được:

$\quad \dfrac{KI}{AM}=\dfrac{KN}{MN}$

$\Rightarrow KI =\dfrac{AM.KN}{MN}$

$\Rightarrow KI =\dfrac{x\cdot \dfrac{a^2}{4\sqrt{\dfrac{3a^2}{4} + x^2}}}{\sqrt{\dfrac{3a^2}{4} + x^2}}$

$\Rightarrow KI = \dfrac{a^2x}{3a^2 + 4x^2}$

$\Rightarrow KI =\dfrac{a^2}{\dfrac{3a^2}{x} + 4x}$

$\Rightarrow KI \leqslant \dfrac{a^2}{2\sqrt{\dfrac{3a^2}{x}\cdot 4x}}$

$\Rightarrow KI \leqslant \dfrac{a}{4\sqrt3}$

Dấu $=$ xảy ra $\Leftrightarrow \dfrac{3a^2}{x}=4x\Leftrightarrow x = \dfrac{a\sqrt3}{2}$

Vậy $x = \dfrac{a\sqrt3}{2}$