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

$\frac{dy}{dx}=\frac{(x-2)+(y-2)}{(x-2)-(y-2)}$  Put  $x-2=h,y-2=k$
$\frac{dk}{dh}=\frac{h+k}{h-k}$ put  $k=vh\Rightarrow v+h\frac{dv}{dh}=\frac{1+v}{1-v}$ 

$\Rightarrow {\tan }^{-1}\left(\frac{y-2}{x-2}\right)=\frac{1}{2}\ln (1+\frac{{(y-2)}^2}{{(x-2)}^2})+\ln (x-2)+c$
   it passes through (3, 2)
$\Rightarrow {\tan }^{-1}0=\frac{1}{2}\ln 1+\ln 1+c\Rightarrow c=0$
   it also passes through (p + 2, 3)
${\tan }^{-1}\left(\frac{1}{p}\right)=\frac{1}{2}\ln (1+\frac{1}{p^2})+\ln p$

 $\Rightarrow 2{\tan }^{-1}\left(\frac{1}{p}\right)=\ln (1+p^2)$