One, (20 points) For the equation in real numbers about $x$
$$
3 x+7 t-2+\frac{2 t x^{2}+7 t^{2}-9}{x-t}=0
$$

there is only one real root. Find the value of the real number $t$.

保留了原文的换行和格式，以下是翻译结果：

One, (20 points) For the equation in real numbers about $x$
$$
3 x+7 t-2+\frac{2 t x^{2}+7 t^{2}-9}{x-t}=0
$$

there is only one real root. Find the value of the real number $t$.
When $t=-\frac{3}{2}$, the root of equation (1) is $-\frac{3}{2}$, and the original equation has no solution.
When $x=t$ is a root of equation (1),
$$
\begin{array}{l}
(2 t+3) t^{2}+(4 t-2) t+2 t-9=0 \\
\Rightarrow t=-\frac{3}{2},-3,1 .
\end{array}
$$

If $t=-3$, then the roots of equation (1) are $-3, -\frac{5}{3}$, and the root of the original equation is $-\frac{5}{3}$, which meets the requirement;

If $t=1$, then the roots of equation (1) are $-\frac{7}{5}, 1$, and the root of the original equation is $-\frac{7}{5}$, which meets the requirement.
When equation (1) has two equal real roots, then
$$
\left\{\begin{array}{l}
2 t+3 \neq 0, \\
\Delta=32 t+112=0
\end{array} \Rightarrow t=-\frac{7}{2},\right.
$$

At this time, the two roots of equation (1) are both 2, and the root of the original equation is 2, which meets the requirement.
In summary, the values of the real number $t$ are $-3, -\frac{7}{2}$, or 1.