If the solution curve of the differential equation $\frac{dy}{dx}=\frac{x+y-4}{x-y}$ passes through the point $(3,2)$ and $(P+2,3),P>0$ then

$$\begin{gathered} \frac{dy}{dx}=\frac{(x-2)+(y-2)}{(x-2)-(y-2)} \\ Put x-2=h,\text{ y-2=k } \\ \frac{dk}{dh}=\frac{h+k}{h-k}met\text{ k=vh}\to \text{v+h}\frac{dv}{dh}=\frac{1+v}{1-v} \\ \Rightarrow \int \frac{1-V}{1+V^2}dv=\int \frac{1}{h}dh \\ \Rightarrow {\tan }^{-1}\left(v\right)=\frac{1}{2}\log (1+v^2)+\log h+c \\ {\tan }^{-1}\left(\frac{y-2}{x-2}\right)=\frac{1}{2}\log (1+\frac{{(y-2)}^2}{{(x-2)}^2})+\log (x-2)+c\text{ \_\_\_\_\_\_ }\left(1\right) \\ (3,2)\Rightarrow 0=0+c\Rightarrow c=0 \end{gathered}$$

Also (1) passes through  $(P+2,3)$

$$\begin{gathered} {\tan }^{-1}\left(\frac{1}{p}\right)=\frac{1}{2}\log (1+\frac{1}{p^2})+\log p \\ 2{\tan }^{-1}\left(\frac{1}{p}\right)=\log (1+p^2) \end{gathered}$$