One. (20 points) Given that $a$, $b$, and $c$ are three distinct real numbers, and the three quadratic equations
\[
\begin{array}{l}
x^{2} + a x + b = 0, \\
x^{2} + b x + c = 0, \\
x^{2} + c x + a = 0
\end{array}
\]

each pair of which has exactly one common root. Find the value of $a^{2} + b^{2} + c^{2}$.
By solving the three equations in pairs, we get
$$
x_{1}=\frac{c-b}{a-b}, x_{2}=\frac{a-c}{b-c}, x_{3}=\frac{b-a}{c-a} \text {. }
$$

Obviously, the three roots $x_{1}, x_{2}, x_{3}$ are distinct, and
$$
x_{1} x_{2} x_{3}=-1 \text {. }
$$

By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1} x_{2}=c, x_{2} x_{3}=a, x_{3} x_{1}=b \\
\Rightarrow a b c=\left(x_{1} x_{2} x_{3}\right)^{2}=1 .
\end{array}
$$

Thus, $x_{1}=-\frac{1}{a}, x_{2}=-\frac{1}{b}, x_{3}=-\frac{1}{c}$.
Also, $x_{1}+x_{2}=-b$, which means
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b}=b \\
\Rightarrow b c+a c=b \\
\Rightarrow a^{2} c+1=a b, b^{2} c+1=b^{2} \\
\Rightarrow a^{2} c=a b-1, b^{2} c=b^{2}-1 .
\end{array}
$$

Similarly, $a b+a c=c$,
$$
\begin{array}{l}
a b+c b=a . \\
\text { Hence } b^{2} a=b c-1, c^{2} a=c^{2}-1, \\
c^{2} b=c a-1, a^{2} b=a^{2}-1 .
\end{array}
$$
(1) + (2) + (3) gives
$$
\begin{array}{l}
a+b+c=2(a b+b c+c a) \\
\Rightarrow(a+b+c)^{2}=2(a+b+c)(a b+b c+c a) \\
\Rightarrow a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \\
\quad=2\left(3 a b c+a^{2} b+b^{2} c+c^{2} a+a^{2} c+c^{2} b+b^{2} a\right) \\
\quad=2\left(a^{2}+b^{2}+c^{2}+a b+b c+c a-3\right) \\
\Rightarrow a^{2}+b^{2}+c^{2}=6 .
\end{array}
$$