11. If the solution set of the inequality $\frac{x^{2}+\left(2 a^{2}+2\right) x-a^{2}+4 a-7}{x^{2}+\left(a^{2}+4 a-5\right) x-a^{2}+4 a-7}<0$ with respect to $x$ is the union of some intervals, and the sum of the lengths of these intervals is not less than 4, find the range of real number $a$.
11. Let the solutions of the equation $x^{2}+\left(2 a^{2}+2\right) x-a^{2}+4 a-7=0$ be $x_{1} 、 x_{2}$, and the solutions of the equation $x^{2}+\left(a^{2}+4 a-5\right) x-a^{2}+4 a-7=0$ be $x_{3} 、 x_{4}$, and $x_{1}<x_{3}<x_{4}<x_{2}$. Since $x_{1}0$, i.e., $x_{3}+x_{4}>x_{1}+x_{2}$. Thus, $a^{2}-4 a+7 \geqslant 4$, yielding $a \leqslant 1$ or $a \geqslant 3$.