\[x^{2} - 4ax + 3a^{2} + 2a - 1 = 0\]
\[D = 16a^{2} - 12a^{2} - 8a + 4 =\]
\[= 4a^{2} - 8a + 4 = (2a - 2)^{2};\ \ \]
\[a \neq 1\]
\[x_{1} = \frac{4a + 2a - 2}{2} = \frac{6a - 2}{2} =\]
\[= 3a - 1\]
\[x_{2} = \frac{4a - 2a + 2}{2} = \frac{2a + 2}{2} =\]
\[= a + 1\]
\[\left\{ \begin{matrix} 3 \leq 3a - 1 \leq 10 \\ 3 \leq a + 1 \leq 10\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 4 \leq 3a \leq 11 \\ 2 \leq a \leq 9\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 1\frac{1}{3} \leq a \leq 3\frac{2}{3} \\ 2 \leq a \leq 9\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[Ответ:\ \left\lbrack 2;3\frac{2}{3} \right\rbrack.\]