Find the point of intersection of the tangents to the curve $y=(x-2)^2-1$ at its points of intersection with the line $x-y=3$

Question

$$x=y+3$$ Then $$y=(y+3-2)^2-1$$ $$y=0,-1$$

So $$x=3,2$$

Then points of intersection of line with parabola are $(3,0)$ and $(2,-1)$

Now $$y=(x-2)^2-1$$ $$\frac{dy}{dx}=2(x-2)(-2)=-4(x-2)$$

The slopes of tangents will be $-4, 0$

Then tangents will be $$y-0=-4(x-3)$$ And $$y+1=0$$ $$y=-1$$

Therefore $$-1=-4(x-3)$$ $$\frac 14 =x-3$$ $$x=\frac{13}{4}$$

So the POI will be $(\frac{13}{4}, -1)$

But then answer is $(\frac 52, -1)$