3. Since the parabola intersects the x-axis at $x=1$, the value of the function $y=a x^{2}+b x+c$ at this point is zero, i.e.,

$$
a+b+c=0
$$

The equation of the given line can be written as

$$
y=5-2 x
$$

The condition for the line to be tangent to the parabola at the point $(x ; y)$ is that the values of the functions and their derivatives are equal:

$$
\left\{\begin{array}{l}
a x^{2}+b x+c=5-2 x \\
2 a x+b=-2
\end{array}\right.
$$

This gives the following system of equations

$$
\left\{\begin{array}{l}
a x^{2}+b x+c=5-2 x \\
2 a x+b=-2 \\
a+b+c=0
\end{array}\right.
$$

From the last two equations, we find

$$
b=-2-2 a x, c=-a-b=-a+2+2 a x
$$

Substituting this into the first equation, we get

$$
a\left(x^{2}-2 x+1\right)=-3, a(x-1)^{2}=-3
$$

According to the problem, the values of $a$ and $x$ must be integers. This is only possible in two cases: $x=0$ or $x=2$. Solving the system, we find two solutions

$$
\begin{aligned}
& x=0 \Rightarrow a=-3, b=-2, c=5 \\
& x=2 \Rightarrow a=-3, b=10, c=-7
\end{aligned}
$$

In the first case, the coordinates of the vertex of the parabola are

$$
x_{b}=-\frac{b}{2 a}=-\frac{1}{3}, y_{s}=a x_{b}^{2}+b x_{b}+c=\frac{16}{3}
$$

And in the second case

$$
x_{6}=-\frac{b}{2 a}=\frac{5}{3}, y_{\sigma}=a x_{\beta}^{2}+b x_{6}+c=\frac{4}{3}
$$
Answer: $\left(-\frac{3}{5} ; \frac{16}{3}\right),\left(\frac{5}{3} ; \frac{4}{3}\right)$.