Lời giải chi tiết:

Bước 1: Giả sử \(A\left( {a;b} \right).\,\,M\left( {3;0} \right)\) là trung điểm AD \( \Rightarrow D\left( {6 - a; - b} \right)\)

Bước 2: Lập 2 phương trình:

\(\begin{array}{l}\left\{ \begin{array}{l}\overrightarrow {AD}  = \left( {6 - 2a; - 2b} \right)\\\overrightarrow {MI}  = \left( {\frac{3}{2};\frac{3}{2}} \right)\end{array} \right.\\\overrightarrow {AD} .\overrightarrow {MI}  = 0 \Rightarrow  - a - b + 3 = 0\,\,\left( 1 \right)\\{S_{\Delta MAI}} = \frac{{{S_{HCN}}}}{8} = \frac{{12}}{8} = \frac{3}{2}\\\left\{ \begin{array}{l}\overrightarrow {MA}  = \left( {a - 3;b} \right)\\\overrightarrow {MI}  = \left( {\frac{3}{2};\frac{3}{2}} \right)\end{array} \right. \Rightarrow \frac{1}{2}\left\| {\begin{array}{*{20}{c}}{a - 3}&b\\{\frac{3}{2}}&{\frac{3}{2}}\end{array}} \right\| = \frac{3}{2}\\ \Rightarrow \left[ \begin{array}{l} - a + b + 1 = 0\,\,\left( 2 \right)\\ - a + b + 5 = 0\,\,\left( {2'} \right)\end{array} \right.\end{array}\)

Giải hệ \(\left\{ \begin{array}{l}\left( 1 \right)\\\left( 2 \right)\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a = 2\\b = 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}A\left( {2;1} \right)\\D\left( {4; - 1} \right)\end{array} \right.\,\,\left( {tm} \right)\)

Giải hệ \(\left\{ \begin{array}{l}\left( 1 \right)\\\left( {2'} \right)\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a = 4\\b =  - 1\end{array} \right. \Rightarrow \left| \begin{array}{l}A\left( {4; - 1} \right)\\D\left( {2;1} \right)\end{array} \right.\,\,\left( {ktm} \right)\)