I tried to solve this problem
$$ \begin{cases} 3u_x+4u_y+5u_z=u+x\\ u(0,0,z)=z^2-1 \end{cases} $$
So, these are the steps I did in order to solve it:
$$ \frac{3}{dx}=\frac{4}{dy}=\frac{5}{dz}=\frac{u+x}{du} $$
$$ \begin{cases} \frac{3}{dx}=\frac{4}{dy}\\ \frac{4}{dy}=\frac{5}{dz}\\ \frac{3}{dx}=\frac{u+x}{du} \end{cases} $$
$$ \begin{cases} 3dy=4dx\\ 4dz=5dy\\ (u+x)dx=3du \end{cases} $$
$$ \begin{cases} 3y+k_1=4x+k_2\\ 4z+k_3=5y+k_4\\ ux+\frac{x^2}{2}+k_5=3u+k_6 \end{cases} $$
$$ \begin{cases} k_1-k_2=4x-3y\\ k_3-k_4=5y-4z\\ ux-3u+\frac{x^2}{2}=k_6-k_5 \end{cases} $$
$$ \begin{cases} \zeta_1:=k_1-k_2=4x-3y\\ \zeta_2:=k_3-k_4=5y-4z\\ u= \frac{\Phi(4x-3y,5y-4z)-\frac{x^2}{2}}{x-3} \end{cases} $$
Now, I tried to sketch the particular solution:
$$ \frac{\Phi(0,-4z)-\frac{x^2}{2}}{x-3}=z^2-1 \quad\rightarrow\quad \Phi(0,-4z)=(z^2-1)(x-3)+\frac{x^2}{2} $$
$$ \begin{cases} \zeta_1=0\\ \zeta_2=-4z \end{cases} $$
$$ \begin{cases} x \text{ undefined}\\ y \text{ undefined}\\ z=-\frac{1}{4}\zeta_2 \end{cases} $$
So $$u=\frac{((-\frac{1}{4}\zeta_2)^2-1)-\frac{x^2}{2}}{x-3} =\ldots$$
and
$$u(x,y,z)=\ldots=\frac{\frac{25}{16}y^2+z^2-\frac{5}{2}yz-1-\frac{x^2}{2}}{x-3}$$
I already noticed that, replacing this solution in the initial value condition, this is wrong. Is it possible that I have been made some "simple" computation errors? Or else, is there some "heavier" concept errors? If so, can someone help me to rebuild the particular solution? Thanks in advance.
You made a mistake in solving $\quad (u+x)dx=3du$
The solution of this first order ODE is not $\quad ux+\frac{x^2}{2}+k_5=3u+k_6 \quad$ because $\int ux\,dx \neq ux$.
The solution is $\quad u=Ce^{x/3}-x-3$ $$(u+x+3)e^{-x/3}=C$$ The general solution of the PDE is : $$u=-x-3+e^{x/3}\Phi\left(4x-3y\:,\:5y-4z \right)$$ I suppose that you can take it from here.