Lời giải chi tiết:

Bước 1 : Giả sử

\(\left\{ \begin{array}{l}B\left( {a;b} \right) \in \left( {{d_1}} \right)\\C\left( {c;d} \right) \in \left( {{d_2}} \right)\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a + b - 3 = 0\\c + d - 9 = 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}b = 3 - a\\d = 9 - c\end{array} \right. \Rightarrow \left\{ \begin{array}{l}B\left( {a;3 - a} \right)\\C\left( {c;9 - c} \right)\end{array} \right.\)

Bước 2 : Lập 2 phương trình :

\(\begin{array}{l}A{B^2} = A{C^2} \Leftrightarrow {\left( {a - 3} \right)^2} + {\left( {1 - a} \right)^2} = {\left( {c - 3} \right)^2} + {\left( {7 - c} \right)^2}\,\,\left( 1 \right)\\\left\{ \begin{array}{l}\overrightarrow {AB}  = \left( {a - 3;1 - a} \right)\\\overrightarrow {AC}  = \left( {c - 3;7 - c} \right)\end{array} \right.\\\overrightarrow {AB} .\overrightarrow {AC}  = 0 \Leftrightarrow \left( {a - 3} \right)\left( {c - 3} \right) + \left( {1 - a} \right)\left( {7 - c} \right) = 0\,\,\left( 2 \right)\end{array}\)

Giải hệ \(\left\{ \begin{array}{l}\left( 1 \right)\\\left( 2 \right)\end{array} \right. \Rightarrow \left[ \begin{array}{l}B\left( {0;3} \right);\,\,C\left( {4;5} \right)\,\,\left( {ktm} \right)\\B\left( {4; - 1} \right);\,\,C\left( {6;3} \right)\,\,\left( {tm} \right)\end{array} \right.\).