7. Let $x_{1}, x_{2}$ be the two real roots of the quadratic equation $x^{2} + a x + a = 2$. Then the maximum value of $\left(x_{1}-2 x_{2}\right)\left(x_{2}-2 x_{1}\right)$ is $\qquad$ .
$$
\begin{array}{l}
\text { II.7. }-\frac{63}{8} . \\
\because \Delta=a^{2}-4(a-2)=(a-2)^{2}+4>0,
\end{array}
$$

Then $x_{1}+x_{2}=-a, x_{1} x_{2}=a-2$.
$$
\begin{array}{l}
\therefore\left(x_{1}-2 x_{2}\right)\left(x_{2}-2 x_{1}\right)=-2 x_{1}^{2}-2 x_{2}^{2}+5 x_{1} x_{2} \\
=-2\left(x_{1}+x_{2}\right)^{2}+9 x_{1} x_{2}=-2 a^{2}+9(a-2) \\
=-2\left(a-\frac{9}{4}\right)^{2}-\frac{63}{8} \leqslant-\frac{63}{8} .
\end{array}
$$

Therefore, when $a=\frac{9}{4}$, $\left(x_{1}-2 x_{2}\right)\left(x_{2}-2 x_{1}\right)$ reaches its maximum value
$$
-\frac{63}{8} \text {. }
$$