Three, (12 points) Let the equation about $x$ be $x^{2}+\left(\frac{a}{x}\right)^{2}-7 x$ $-\frac{7 a}{x}+2 a+12=0$ and suppose it has equal roots. Find the value of $a$.

 untranslated portion: 

设关于 $x$ 的方程 $x^{2}+\left(\frac{a}{x}\right)^{2}-7 x$ $-\frac{7 a}{x}+2 a+12=0$ 有相等鳫根. 求 $a$ 的值.

translated portion:

Let the equation about $x$ be $x^{2}+\left(\frac{a}{x}\right)^{2}-7 x$ $-\frac{7 a}{x}+2 a+12=0$ and suppose it has equal roots. Find the value of $a$.
Three, transform the original equation into
$$
\left(x+\frac{a}{x}-4\right)\left(x+\frac{a}{x}-3\right)=0 .
$$

Then we have $x+\frac{a}{x}-4=0$ or $x+\frac{a}{x}-3=0$,
i.e., $\square$
$$
\begin{array}{l}
x^{2}-4 x+a=0, \\
x^{2}-3 x+a=0 .
\end{array}
$$

If (1) has a repeated root, we get $a=4$.
If (2) has a repeated root, we get $a=2 \frac{1}{4}$,
If (1) and (2) have a repeated root, we get $x=0$, which is impossible.
Therefore, the value of $a$ is 4 or $2 \frac{1}{4}$.