26 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 $a c+b c-c^{2} \leqslant \lambda\left(a^{2}+b^{2}+3 c^{2}+2 a b-4 b c\right)$.
26. It is known that $a^{2}+b^{2}+3 c^{2}+2 a b-4 b c=(a+b-c)^{2}+2 c^{2}+2 a c-2 b c=$ $(a+b-c)^{2}+2 c(a+c-b)$. 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 $a \geqslant \frac{1}{3}(b+c)$, it follows that $a \geqslant \frac{1}{4}(a+b-c)+\frac{c}{2}$. Therefore, $a+c-$ $b=2 a-(a+b-c) \geqslant-\frac{1}{2}(a+b-c)+c$. From this, we know that $I \geqslant \frac{a+b-c}{2 c}-$ $\frac{\frac{1}{2}(a+b-c)}{a+b-c}+\frac{c}{a+b-c}=-\frac{1}{2}+\frac{a+b-c}{2 c}+\frac{c}{a+b-c} \geqslant-\frac{1}{2}+2 \sqrt{\frac{1}{2}}=$ $\sqrt{2}-\frac{1}{2}$. 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, $\lambda \geqslant \frac{2 \sqrt{2}+1}{7}$. 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 $a c+b c-$ $c^{2}=\sqrt{2}, a^{2}+b^{2}+3 c^{2}+2 a b-4 b c=4-\sqrt{2}$, so $\frac{1}{2 I}=\frac{\sqrt{2}}{4-\sqrt{2}}=\frac{2 \sqrt{2}+1}{7}$. Hence, $\lambda=\frac{2 \sqrt{2}+1}{7}$.