I am trying to solve the following pde:
$$(x+z)u_x + (y+z)u_y + (x+y)u_z = 0$$
My attempt is writing the following characteristic equations
$$\frac{dx}{x+z}= \frac{dy}{y+z} = \frac{dz}{x+y}= \frac{du}{0} $$
and then $$ \frac{dy -dz}{z-x}= \frac{dz-dx}{y-z}$$
From here I am stuck, I tried using the relationship of the $d$ operator to get $$ \frac{d(y -z)}{z-x}= \frac{d(z-x)}{y-z}$$ and then integrated the expression above by substituting $y-z$ with $t$ but it feels like its not right. What approach could I use to solve this pde
$$(x+z)u_x + (y+z)u_y + (x+y)u_z = 0$$
You correctly wrote the Charpit-Lagrange ODEs : $$\frac{dx}{x+z}= \frac{dy}{y+z} = \frac{dz}{x+y}= \frac{du}{0} $$
Also you correctly found one of the linear combinaison of ODEs : $\quad \frac{dy -dz}{z-x}= \frac{dz-dx}{y-z}$
A first characteristic equation comes from solving it :
$(y-z)(dy-dz)=(z-x)(dz-dx)\quad\implies\quad (y-z)^2=(z-x)^2+c_1$. After simplification : $$(y-x)(x+y-2z)=c_1$$
Another linear combination of the three ODEs leads to : $$\frac{dx}{x+z}= \frac{dy}{y+z} = \frac{dz}{x+y}= \frac{dx+dy-2dz}{(x+z)+(y+z)-2(x+y)} =\frac{dx+dy-2dz}{0}$$ This implies $(dx+dy-2dz)=0$ and a second characteristic equation : $$x+y-2z=c_2$$ The general solution of the PDE is : $$u(x,t,z)=F\big((y-x)(x+y-2z)\:\:,\:\:(x+y-2z) \big)$$ $F$ is an arbitrary function of two variables.