II. (25 points) Let $a$, $b$, and $c$ be three distinct real numbers, and $c \neq 1$. It is known that the equations $x^{2} + a x + 1 = 0$ and $x^{2} + b x + c = 0$ have a common root, and the equations $x^{2} + x + a = 0$ and $x^{2} + c x + b = 0$ also have a common root. Find the value of $a + b + c$.
Let the common root of the first two equations be $x_{1}$, then
$$
\begin{array}{l}
x_{1}^{2}+a x_{1}+1=0, \\
x_{1}^{2}+b x_{1}+c=0 .
\end{array}
$$
(2)
(1) - (2) gives $(a-b) x_{1}+(1-c)=0$.
$$
\because a \neq b, \quad \therefore x_{1}=\frac{c-1}{a-b} \text {. }
$$

Similarly, the common root of the last two equations is $x_{2}=\frac{a-b}{c-1}$.
$$
\therefore x_{2}=\frac{1}{x_{1}} \text {. }
$$

By Vieta's formulas, the other root of the equation $x^{2}+a x+1=0$ is $\frac{1}{x_{1}}$. Therefore, $x_{2}$ is the common root of the equations $x^{2}+a x+1=0$ and $x^{2}+x+a=0$. Thus, we have
$$
x_{2}^{2}+a x_{2}+1=0, \quad x_{2}^{2}+x_{2}+a=0 .
$$

Subtracting these gives $(a-1)\left(x_{2}-1\right)=0$.
When $a=1$, these two equations have no real roots, so $a \neq 1$. Hence, we have $x_{2}=1$, and thus $x_{1}=1$. Therefore, $a=-2, b+c=-1$.
$$
\therefore a+b+c=-3 .
$$