\[\begin{array}{l}{\text{I}}\\{y^2} - \left( {2a + 1} \right)y + {a^2} + a - 2 = 0 \Leftrightarrow \left( {y - \left( {a - 1} \right)} \right) \cdot \left( {y - \left( {a + 2} \right)} \right) = 0\\{\text{II}}\\\sqrt {{{\left( {x - a} \right)}^2} + {y^2}} + \sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - 3} \right)}^2}} = 3 \Leftrightarrow \left\{ \begin{array}{l}x = a\\\left| y \right| + \left| {y - 3} \right| = 3\end{array} \right.\end{array}\]

\[\begin{array}{l}{y^2} - \left( {2a + 1} \right)y + {a^2} + a - 2 = 0 \Leftrightarrow \\\left( {y - \left( {a - 1} \right)} \right) \cdot \left( {y - \left( {a + 2} \right)} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}y = a - 1\\y = a + 2\end{array} \right.\end{array}\]
\[\begin{array}{l}\sqrt {{{\left( {x - a} \right)}^2} + {y^2}} + \sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - 3} \right)}^2}} \ge \\\sqrt {{y^2}} + \sqrt {{{\left( {y - 3} \right)}^2}} = \left| y \right| + \left| {y - 3} \right| \ge 3 \Rightarrow \\{\left( {x - a} \right)^2} = 0 \Leftrightarrow x = a\end{array}\]
\[\begin{array}{l}{\text{Тогда второе уравнение перепишется в виде}}\\\left| y \right| + \left| {y - 3} \right| = 3 \Leftrightarrow 0 \le y \le 3.\\{\text{Исходная система в результате равносильна системе }}\\\left\{ \begin{array}{l}x = a\\\left[ \begin{array}{l}y = a - 1\\y = a + 2\end{array} \right.\\0 \le y \le 3\end{array} \right.\end{array}\]
\[\begin{array}{l}0 \le a - 1 \le 3 \Leftrightarrow 1 \le a \le 4\\0 \le a + 2 \le 3 \Leftrightarrow - 2 \le a \le 1\\a = 1{\text{ не подходит}}{\text{, в этом случае система}}\\{\text{имеет два решения }}\left( {1;0} \right){\text{ и }}\left( {1;3} \right).\end{array}\]

\[a \in \left[ { - 2;1} \right) \cup \left( {1;4} \right]\]