Given that $a$, $b$, and $c$ are three distinct real numbers. If in the quadratic equations
$$
\begin{array}{l}
x^{2}+a x+b=0, \\
x^{2}+b x+c=0, \\
x^{2}+c x+a=0
\end{array}
$$

any two of these equations have exactly one common root, find the value of $a^{2}+$ $b^{2}+c^{2}$.
Let the equations (1) and (3) have only one common root \( x_{1} \), equations (1) and (2) have only one common root \( x_{2} \), and equations (2) and (3) have only one common root \( x_{3} \). Therefore, the roots of equation (1) are \( x_{1} \) and \( x_{2} \), the roots of equation (2) are \( x_{2} \) and \( x_{3} \), and the roots of equation (3) are \( x_{3} \) and \( x_{1} \).

Since \( x_{1} \) is a root of equations (1) and (3), we have
\[
\left\{\begin{array}{l}
x_{1}^{2} + a x_{1} + b = 0, \\
x_{1}^{2} + c x_{1} + a = 0
\end{array} \Rightarrow x_{1} = \frac{a - b}{a - c} .\right.
\]

Similarly, \( x_{2} = \frac{b - c}{b - a} \) and \( x_{3} = \frac{c - a}{c - b} \).
Thus, \( x_{1} x_{2} x_{3} = -1 \).
Clearly, \( x_{1} \), \( x_{2} \), and \( x_{3} \) are distinct. Otherwise, assume \( x_{2} = x_{3} \).
\[
\begin{array}{l}
\text{Hence } \frac{b - c}{b - a} = \frac{c - a}{c - b} \\
\Rightarrow a^{2} + b^{2} + c^{2} - b c - c a - a b = 0 \\
\Rightarrow (b - c)^{2} + (c - a)^{2} + (a - b)^{2} = 0 \\
\Rightarrow a = b = c,
\end{array}
\]
which contradicts the given conditions.
Thus, \( x_{1} \), \( x_{2} \), and \( x_{3} \) are distinct.
By Vieta's formulas, we have
\[
\begin{array}{l}
a = -x_{1} - x_{2} = x_{3} x_{1}, \\
b = -x_{2} - x_{3} = x_{1} x_{2}, \\
c = -x_{3} - x_{1} = x_{2} x_{3}. \\
\text{From the above three equations, we get} \\
-2(x_{1} + x_{2} + x_{3}) = x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2}, \text{ (4) } \\
x_{1}(1 + x_{3}) x_{2}(1 + x_{1}) x_{3}(1 + x_{2}) \\
= (-x_{2})(-x_{3})(-x_{1}). \\
\text{Noting that } x_{1} x_{2} x_{3} = -1. \\
\text{Thus } (1 + x_{3})(1 + x_{1})(1 + x_{2}) = -1 \\
\Rightarrow x_{1} + x_{2} + x_{3} + x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -1. \text{ (5) } \\
\text{From equations (4) and (5), we get} \\
\left\{\begin{array}{l}
x_{1} + x_{2} + x_{3} = 1, \\
x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -2
\end{array}\right. \\
\Rightarrow \left\{\begin{array}{l}
a + b + c = -2, \\
b c + c a + a b = -1 \\
a b c = 1
\end{array}\right., \\
\Rightarrow a^{2} + b^{2} + c^{2} \\
= (a + b + c)^{2} - 2(b c + c a + a b) \\
= (-2)^{2} - 2(-1) = 6. \\
\end{array}
\]

Let's explore further.
From the above solution, we also obtained
\[
\left\{\begin{array}{l}
x_{1} + x_{2} + x_{3} = 1, \\
x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -2 \\
x_{1} x_{2} x_{3} = -1.
\end{array}\right.
\]

Therefore, \( a \), \( b \), and \( c \) are the roots of the equation \( x^{3} + 2 x^{2} - x - 1 = 0 \); simultaneously, \( x_{1} \), \( x_{2} \), and \( x_{3} \) are the roots of the equation \( x^{3} - x^{2} - 2 x + 1 = 0 \). If we substitute \( x \rightarrow -\frac{1}{x} \) in the equation \( x^{3} + 2 x^{2} - x - 1 = 0 \), we get the equation \( x^{3} - x^{2} - 2 x + 1 = 0 \). This is an interesting conclusion.

If we use conclusion (6), we can derive many symmetric expressions involving \( a \), \( b \), and \( c \). However, if we need to find the values of \( a^{2} b + b^{2} c + c^{2} a \) and \( a b^{2} + b c^{2} + c a^{2} \), the above solution is not straightforward. We will use a different approach to find these values.
Clearly, the six roots of these three equations must be the roots of the equation \( \left(x^{2} + a x + b\right)\left(x^{2} + b x + c\right)\left(x^{2} + c x + a\right) = 0 \),

which is \( x^{6} + (a + b + c) x^{5} + (a + b + c + b c + c a + a b) x^{4} + (a^{2} + b^{2} + c^{2} + b c + c a + a b + a b c) x^{3} + (a^{2} b + b^{2} c + c^{2} a + b c + c a + a b) x^{2} + (a b^{2} + b c^{2} + c a^{2}) x + a b c = 0 \).

Let \( s_{1} = x_{1} + x_{2} + x_{3} \),
\( s_{2} = x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} \), \( s_{3} = x_{1} x_{2} x_{3} \). By the relationship between the roots and coefficients, and according to conclusion (6), we have
\[
\begin{array}{l}
\left\{\begin{array}{l}
s_{1} = x_{1} + x_{2} + x_{3} = 1, \\
s_{2} = x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -2, \\
s_{3} = x_{1} x_{2} x_{3} = -1.
\end{array}\right. \\
\left\{\begin{array}{l}
a + b + c = -2 s_{1}, \\
a + b + c + b c + c a + a b = s_{1}^{2} + 2 s_{2}, \\
a^{2} + b^{2} + c^{2} + b c + c a + a b + a b c \\
= -2 s_{1} s_{2} - 2 s_{3}, \\
a^{2} b + b^{2} c + c^{2} a + b c + c a + a b \\
= s_{2}^{2} + 2 s_{1} s_{3}, \\
a b^{2} + b c^{2} + c a^{2} = -2 s_{2} s_{3}, \\
a b c = s_{3}^{2}.
\end{array}\right.
\end{array}
\]

From equations (7) and (8), we get
\[
\begin{array}{l}
a^{2} b + b^{2} c + c^{2} a = s_{2}^{2} + 2 s_{1} s_{3} - (b c + c a + a b) \\
= (-2)^{2} + 2 \times 1(-1) - (-1) = 3, \\
a b^{2} + b c^{2} + c a^{2} = -2 s_{2} s_{3} \\
= -2(-2)(-1) = -4.
\end{array}
\]