Если \(a-3 = 0\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,a = 3,\) то неравенство примет вид: \(0 \cdot {x^2}-6x + 9-6 \ge 0\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x \le \dfrac{1}{2},\) то есть \(a = 3\) не подходит.
Если \(a \ne 3\), то исходное неравенство не выполняется ни для одного значения x в случае выполнения следующих условий:
\(\left\{ {\begin{array}{*{20}{c}}{a-3 < 0,\,}\\{D < 0\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a < 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{4{a^2}-4\left( {a-3} \right)\left( {3a-6} \right) < 0}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a < 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{2{a^2}-15a + 18 > 0}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a\, \in \,\left( {-\infty ;3} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;\dfrac{3}{2}} \right) \cup \left( {6;\infty } \right)\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,a\, \in \,\left( {-\infty ;\dfrac{3}{2}} \right).} \right.\)
Следовательно, неравенство не выполняется ни для одного значения x, при \(a\, \in \,\left( {-\infty ;\dfrac{3}{2}} \right)\).
ОТВЕТ: \(\left( {-\infty ;\,\dfrac{3}{2}} \right).\)