c Lu
$$
\begin{array}{l}
30-6(a+b+c)+(a b+b c+c a)=0, \\
81-9(a+b+c)+(a b+b c+c a)=0 .
\end{array}
$$

Find $a, b, c$.
Three, it is known that 6 and 9 are the roots of the quadratic equation
$$
x^{2}-(a+b+c) x+(a b+b c+c a)=0
$$

Thus, we have
$$
\left\{\begin{array}{l}
a+b+c=6+9=15, \\
a b+b c+c a=6 \times 9=54 .
\end{array}\right.
$$

Since $a>b>c$, it follows that
$$
15=a+b+c>3 c,
$$

Solving this, we get $c<5$. If $c \geq 5$, then $a+b \leq 10$, which contradicts $a>b>c$. Therefore, $c<5$.

If $c=4$, then $a+b=11$ and $a b=38$. Solving the equation $t^{2}-11 t+38=0$ yields no real roots, which is a contradiction.

If $c=3$, then $a+b=12$ and $a b=45$. Solving the equation $t^{2}-12 t+45=0$ yields no real roots, which is a contradiction.

If $c=2$, then $a+b=13$ and $a b=50$. Solving the equation $t^{2}-13 t+50=0$ yields no real roots, which is a contradiction.

If $c=1$, then $a+b=14$ and $a b=53$. Solving the equation $t^{2}-14 t+40=0$ yields the roots $a=10$ and $b=4$, which satisfies $a>b>c$.

If $c<1$, then $a+b>14$ and $a b>54$, which contradicts $a>b>c$.

In summary, the only solution is $a=10, b=4, c=1$.
（Provided by Luo Zengru, West Normal University）