9.62 Find the smallest positive number $\lambda$, such that for any triangle with side lengths $a, b, c$, if $a \geqslant \frac{b+c}{3}$, then
$$a c+b c-c^{2} \leqslant \lambda\left(a^{2}+b^{2}+3 c^{2}+2 a b-4 b c\right) .$$
[Solution] It is easy to know that
$$\begin{aligned}
a^{2}+b^{2}+3 c^{2}+2 a b-4 b c & =(a+b-c)^{2}+2 c^{2}+2 a c-2 b c \\
& =(a+b-c)^{2}+2 c(a+c-b) .
\end{aligned}$$

Let \( I=\frac{(a+b-c)^{2}+2 c(a+c-b)}{2 c(a+b-c)}=\frac{a+b-c}{2 c}+\frac{a+c-b}{a+b-c} \).
Since \( a \geqslant \frac{1}{3}(b+c) \), we have \( a \geqslant \frac{1}{4}(a+b-c)+\frac{c}{2} \), thus
$$a+c-b=2 a-(a+b-c) \geqslant-\frac{1}{2}(a+b-c)+c .$$

From this, we know
$$\begin{array}{l}
\quad I \geqslant \frac{a+b-c}{2 c}-\frac{\frac{1}{2}(a+b-c)}{a+b-c}+\frac{c}{a+b-c} \\
\quad=-\frac{1}{2}+\frac{a+b-c}{2 c}+\frac{c}{a+b-c} \\
\quad \geqslant-\frac{1}{2}+2 \sqrt{\frac{1}{2}}=\sqrt{2}-\frac{1}{2} \\
\text { i.e., } \quad \frac{a c+b c-c^{2}}{a^{2}+b^{2}+3 c^{2}+2 a b-4 b c}=\frac{1}{2 I} \leqslant \frac{1}{2 \sqrt{2}-1}=\frac{2 \sqrt{2}+1}{7} .
\end{array}$$

Thus, the required \( \lambda \leqslant \frac{2 \sqrt{2}+1}{7} \).
On the other hand, when \( a=\frac{\sqrt{2}}{4}+\frac{1}{2}, b=\frac{3 \sqrt{2}}{4}+\frac{1}{2}, c=1 \), then
$$\begin{array}{l}
a c+b c-c^{2}=\sqrt{2} \\
a^{2}+b^{2}+3 c^{2}+2 a b-4 b c=4-\sqrt{2}
\end{array}$$

So \( \frac{1}{2 I}=\frac{\sqrt{2}}{4-\sqrt{2}}=\frac{\sqrt{2}(4+\sqrt{2})}{14}=\frac{2 \sqrt{2}+1}{7} \).
Thus, the required \( \lambda \geqslant \frac{2 \sqrt{2}+1}{7} \).
In summary, \( \lambda=\frac{2 \sqrt{2}+1}{7} \).