Three, given the parabola $y=a x^{2}-(a-1) x+a-1$ - the line $y=1-2 x$ has at least one intersection point that is a lattice point (in the Cartesian coordinate system, a point where both the x and y coordinates are integers). Determine the integer value of $a$, and find the coordinates of the intersection point (lattice point) at this time.
Three, for the parabola $y=a x^{2}-(a-1) x+a-1$ and the line $y=1-2 x$, the x-coordinates of their intersection points are the real solutions of the equation $a x^{2}-(a-1) x+(a-1)=1-2 x$, which is
$$
a x^{2}-(a-3) x+a-2=0
$$

The condition for equation (1) to have real solutions is
$$
\Delta=(a-3)^{2}-4 a(a-2)>0 \text {. }
$$

which simplifies to $3 a^{2}-2 a-9 \leqslant 0$.
Solving this, we get $\frac{1-2 \sqrt{7}}{3} \leqslant a \leqslant \frac{1+2 \sqrt{7}}{3}$.
Since $a$ is an integer, $a=-1,0,1, 2$.
When $a=-1$, from equation (1) we get $x=1$ or 3, substituting into $y=1-2 x$, we get $y=-1$ or -5. Thus, the intersection points are $(1,-1)$ and $(3,-5)$.
When $a=0$, equation (1) has no integer solutions.
When $a=1$, equation (1) has no integer solutions.
When $a=2$, from equation (1) we get $x=0$ or $x=-\frac{1}{2}$ (discard). Substituting into $y=1-2 x$, we get $y=1$. Thus, the intersection point is $(0,1)$.