Example 6 Find all positive integer triples $(a, b, c)$ such that the roots of the following three quadratic equations in $x$
$$
\begin{array}{l}
x^{2}-3 a x+2 b=0, \\
x^{2}-3 b x+2 c=0, \\
x^{2}-3 c x+2 a=0
\end{array}
$$

are all positive integers.
Let the roots of the equation $x^{2}-3 a x+2 b=0$ be $x_{1}, x_{2}$, with $x_{1} \leqslant x_{2}$; the roots of $x^{2}-3 b x+2 c=0$ be $x_{3}, x_{4}$, with $x_{3} \leqslant x_{4}$; and the roots of $x^{2}-3 c x+2 a=0$ be $x_{5}, x_{6}$, with $x_{5} \leqslant x_{6}$. By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}=3 a, x_{1} x_{2}=2 b ; \\
x_{3}+x_{4}=3 b, x_{3} x_{4}=2 c ; \\
x_{5}+x_{6}=3 c, x_{5} x_{6}=2 a .
\end{array}
$$

Thus, we have
$$
\begin{array}{l}
3\left(x_{1} x_{2}+x_{3} x_{4}+x_{5} x_{6}\right)=6(a+b+c) \\
=2\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}\right) . \\
\text { Therefore, }\left(9 x_{1} x_{2}-6 x_{1}-6 x_{2}\right)+\left(9 x_{3} x_{4}-\right. \\
\left.6 x_{3}-6 x_{4}\right)+\left(9 x_{5} x_{6}-6 x_{5}-6 x_{6}\right)=0,
\end{array}
$$

which simplifies to
$$
\left(3 x_{1}-2\right)\left(3 x_{2}-2\right)+\left(3 x_{3}-2\right)\left(3 x_{4}-2\right)+\left(3 x_{5}-2\right)\left(3 x_{6}-2\right)=12.
$$

Since $x_{1}, x_{2}$ are positive integers, from $x_{1} x_{2}=2 b$ we know that at least one of $x_{1}, x_{2}$ is even. Therefore,
$$
\left(3 x_{1}-2\right)\left(3 x_{2}-2\right) \geqslant 4.
$$

Similarly, we have
$$
\begin{array}{l}
\left(3 x_{3}-2\right)\left(3 x_{4}-2\right) \geqslant 4 ; \\
\left(3 x_{5}-2\right)\left(3 x_{6}-2\right) \geqslant 4.
\end{array}
$$

Combining with the above equation, we get
$$
\begin{array}{l}
\left(3 x_{1}-2\right)\left(3 x_{2}-2\right)=\left(3 x_{3}-2\right)\left(3 x_{4}-2\right) \\
=\left(3 x_{5}-2\right)\left(3 x_{6}-2\right)=4.
\end{array}
$$

Thus, $x_{1}=x_{3}=x_{5}=1, x_{2}=x_{4}=x_{6}=2$.
Therefore, $a=b=c=1$.
When $a=b=c=1$, the three equations are all $x^{2}-3 x+2=0$, with positive integer roots 1 and 2.
Hence, the required positive integer solution is $(1,1,1)$.