3. In the Cartesian coordinate system, find the coordinates of the point that satisfies the following two conditions simultaneously:
(1) The line $y=-2 x+3$ passes through such a point;
(2) Regardless of the value of $m$, the parabola
$$
y=m x^{2}+\left(m-\frac{2}{3}\right) x-\left(2 m-\frac{3}{8}\right)
$$
3. From (2) we know $m \neq 0$.

Let the point $\left(x_{0}, y_{0}\right)$ satisfy (1) and (2), then
$$
y_{0}=-2 x_{0}+3 \text {, }
$$

and for any non-zero real number $m$, we have
$$
y_{0} \neq m x_{0}^{2}+\left(m-\frac{2}{3}\right) x_{0}-\left(2 m-\frac{3}{8}\right) .
$$

Substituting equation (1) into equation (2) and simplifying, we get
$$
\left(x_{0}-1\right)\left(x_{0}+2\right) m \neq-\frac{4}{3} x_{0}+\frac{21}{8} \text {. }
$$

Therefore, $x_{0}=1,-2$ or $\frac{63}{32}$. Substituting into equation (1) yields the coordinates of the points that satisfy both conditions (1) and (2) as
$$
(1,1),(-2,7),\left(\frac{63}{32},-\frac{15}{16}\right) \text {. }
$$