One. (20 points) Given three different real numbers $a, b, c$ satisfying $a-b+c=3$, the equations
$$
x^{2}+a x+1=0 \text { and } x^{2}+b x+c=0
$$

have one common real root, and the equations
$$
x^{2}+x+a=0 \text { and } x^{2}+c x+b=0
$$

also have one common real root. Find the values of $a, b, c$.
I. Sequentially number the four equations given in the problem as (1), (2), (3), and (4).
Let $x_{1}$ be a common real root of equations (1) and (2).
(1) - (2) gives $x_{1}=\frac{c-1}{a-b}$.
Let $x_{2}$ be a common real root of equations (3) and (4).
(3) - (4) gives $x_{2}=\frac{a-b}{c-1}$.
Thus, $x_{1} x_{2}=1$.
Since the product of the roots of equation (1) is 1, $x_{2}$ is also a root of equation (1).
Therefore, $x_{2}^{2}+a x_{2}+1=0$.
Also, $x_{2}^{2}+x_{2}+a=0$, subtracting these two equations gives $(a-1) x_{2}=a-1$.
If $a=1$, then equation (1) has no real roots.
Therefore, $a \neq 1$. Hence, $x_{2}=1$.
Thus, $a=-2, b+c=-1$.
Also, $a-b+c=3$, solving these gives $b=-3, c=2$.
Therefore, the solution is $a=-2, b=-3, c=2$.