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
$$
a c+b c-c^{2} \leqslant \lambda\left(a^{2}+b^{2}+3 c^{2}+2 a b-4 b c\right) .
$$
\[ \text{Solution: It is easy to see that} \]
\[ 
\begin{array}{l} 
a^{2}+b^{2}+3 c^{2}+2 a b-4 b c=(a+b-c)^{2}+2 c^{2}+2 a c-2 b c \\
=(a+b-c)^{2}+2 c(a+c-b) . \\
\text{Let } \quad I=\frac{(a+b-c)^{2}+2 c(a+c-b)}{2 c(a+b-c)}=\frac{a+b-c}{2 c}+\frac{a+c-b}{a+b-c} .
\end{array}
\]

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
\[ 
\begin{aligned}
I & \geqslant \frac{a+b-c}{2 c}-\frac{\frac{1}{2}(a+b-c)}{a+b-c}+\frac{c}{a+b-c} \\
& =-\frac{1}{2}+\frac{a+b-c}{2 c}+\frac{c}{a+b-c} \\
& \geqslant-\frac{1}{2}+2 \sqrt{\frac{1}{2}}=\sqrt{2}-\frac{1}{2}
\end{aligned}
\]

Thus,
\[ 
\frac{a c+b c-c^{2}}{a^{2}+b^{2}+3 c^{2}+2 a b-4 b c}=\frac{1}{2 I} \leqslant \frac{1}{2 \sqrt{2}-1}=\frac{2 \sqrt{2}+1}{7}.
\]
Therefore, 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, \(\lambda=\frac{2 \sqrt{2}+1}{7}\).