The renormalisation flow equation is equivalent to

$$\begin{cases}\frac{d}{d\lambda}p_{1}=-\frac{p_{2}+p_{3}}{4p_{1}p_{2}p_{3}},\\ \frac{d}{d\lambda}p_{2}=-\frac{p_{1}+p_{3}}{4p_{1}p_{2}p_{3}},\\ \frac{d}{d\lambda}p_{3}=-\frac{p_{1}+p_{2}}{4p_{1}p_{2}p_{3}}.\end{cases}$$

Summing over the three equations,

$$\frac{d}{d\lambda}(p_{1}+p_{2}+p_{3})=-\frac{p_{1}+p_{2}+p_{3}}{2p_{1}p_{2}p_{3}},$$

and taking the differences give

$$\frac{d}{d\lambda}(p_{1}-p_{2})=\frac{p_{1}-p_{2}}{4p_{1}p_{2}p_{3}},\quad\frac{d}{d\lambda}(p_{1}-p_{3})=\frac{p_{1}-p_{3}}{4p_{1}p_{2}p_{3}}.$$

Without loss of generality $p_{1}\geq p_{2}\geq p_{3}$, then

$$d\log(p_{1}+p_{2}+p_{3})=-2d\log(p_{1}-p_{2})=-2\log(p_{1}-p_{3})=-\frac{d\lambda}{2p_{1}p_{2}p_{3}}.$$

In the degenerate case where $p_{1}=p_{2}$ say, it is understood that $p_{1}=p_{2}$ identically. Denote $p_{1}+p_{2}+p_{3}=3t^{2/3}$ for some new parameter $t$, then after integration

$$p_{1}-p_{2}=K_{1}t^{-1/3},\quad p_{1}-p_{3}=K_{2}t^{-1/3},$$

for some constants $0\leq K_{1}\leq K_{2}$. Rewriting these equations give

$$\begin{cases}p_{1}=t^{2/3}+\frac{1}{3}(K_{1}+K_{2})t^{-1/3},\\ p_{2}=t^{2/3}+\frac{1}{3}(K_{2}-2K_{1})t^{-1/3},\\ p_{3}=t^{2/3}+\frac{1}{3}(K_{1}-2K_{2})t^{-1/3}.\end{cases}$$

Now

$$d\lambda=-\frac{4}{3}p_{1}p_{2}p_{3}\frac{dt}{t}=\frac{-4}{3t^{2}}(t+\frac{1}{3}(K_{1}+K_{2}))(t+\frac{1}{3}(K_{1}-2K_{2}))(t+\frac{1}{3}(K_{2}-2K_{1}))dt.$$

Increasing $\lambda$ corresponds to decreasing $t$. We integrate to obtain

$$\lambda=-\frac{2}{3}t^{2}+\frac{4}{9}(K_{1}^{2}+K_{2}^{2}-K_{1}K_{2})\log t+\frac{4}{81}(K_{1}+K_{2})(K_{2}-2K_{1})(K_{1}-2K_{2})\frac{1}{t}+K_{3},$$

where $K_{3}$ is an integration constant. ∎