Example 3. Given positive integers $a, b, c$ satisfying $a>b>c$, and $\quad\left\{\begin{array}{l}36-6(a+b+c)+(a b+b c+c a)=0, \\ 81-9(a+b+c)+(a b+b c+c a)=0 .\end{array}\right.$ Find the values of $a, b, c$.
The problem statement indicates that 6 and 9 are the roots of the quadratic equation in $x$: 
$$x^{2}-(a+b+c) x+(a b+b c+c a)=0.$$
By Vieta's formulas, we get
$$
\left\{\begin{array}{l}
a+b+c=6+9=15 . \\
a b+b c+i a=6 \times 9=54 .
\end{array}\right.
$$

Given $a>b>1$, we have $1.5=a+b+c>3 c$, which implies $c<5$.
$\therefore c$ can only take the values $1,2,3,4$.
(1) Substituting $c=1$ into (1), we get
$$
\left\{\begin{array}{l}
a+b=14, \\
a b=40 .
\end{array}\right.
$$

Solving this, we get $a=10, b=4$.
(2) Substituting $c=2, 3, 4$ into (1), we get
$$
\left\{\begin{array}{l}
a+b=13, \\
a b=28 ;
\end{array} \left\{\begin{array}{l}
a+b=12, \\
a b=18 ;
\end{array} \left\{\begin{array}{l}
a+b=11, \\
a b=10 .
\end{array}\right.\right.\right.
$$

It is easy to see that none of these have positive integer solutions.
In conclusion, the only solution is $a=10, b=4, c=1$.