Введём функцию \(f\left( x \right) = {x^2}-6ax + 2-2a + 9{a^2}\) графиком которой является парабола ветвями вверх.
Для того чтобы уравнение имело два различных корня, каждый из которых больше 3, необходимо выполнение условий (см. рис.):
\(\left\{ {\begin{array}{*{20}{c}}{D > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x_B} = -\dfrac{b}{{2a}} > 3,}\\{f\left( 3 \right) > 0\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{36{a^2}-4\left( {2-2a + 9{a^2}} \right) > 0,}\\{-\dfrac{{-6a}}{2} > 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{9-18a + 2-2a + 9{a^2} > 0\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\)
\( \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{8a-8 > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a > 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{9{a^2}-20a + 11 > 0\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a > 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a > 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;1} \right) \cup \left( {\dfrac{{11}}{9};\infty } \right)}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,a\, \in \,\left( {\dfrac{{11}}{9};\infty } \right).\)
ОТВЕТ: \(\left( {\dfrac{{11}}{9};\,\infty } \right).\)