\(\left( {{x^2}-2x + 3} \right)\left( {{y^2} + 6y + 12} \right) = 6\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left( {{{\left( {x-1} \right)}^2} + 2} \right)\left( {{{\left( {y + 3} \right)}^2} + 3} \right) = 6.\)
Так как \({\left( {x-1} \right)^2} + 2 \ge 2\) и \(\,{\left( {y + 3} \right)^2} + 3 \ge 3\), то \(\left( {{{\left( {x-1} \right)}^2} + 2} \right)\left( {{{\left( {y + 3} \right)}^2} + 3} \right) \ge 6\).
Поэтому исходное уравнение будет иметь решение, если:
\(\left\{ \begin{array}{l}{\left( {x-1} \right)^2} + 2 = 2,\\{\left( {y + 3} \right)^2} + 3 = 3\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 1,\\y = -3.\end{array} \right.\)
Ответ: \(\left( {1;\,-3} \right).\)