Geometric interpretation and solution of Lagrange-Charpit equations What is the geometric meaning of the Lagrange-Charpit equation, $$P\dfrac{\partial z}{\partial x} + Q \dfrac {\partial z}{\partial y}= R \\Pp+ Qq= R$$ where $P$,$Q$,and $R$  are functions of $x\,$,$\,y\,$ and $z$. How do we arrive at the solution of the same? I've got a feeling that it has something to do with direction cosines and stuff. But I just cant relate them with the equation. We are being taught just to write the auxiliary equation, $$\dfrac{dx}{P}=\dfrac{dy}{Q}=\dfrac{dz}{R}$$ group, and integrate or find two set of "multipliers" and finally write $\phi (u,v)=0$. I've got no clue, what's happening behind the curtains.
P.S: I'm relatively new to the subject
 A: First Interpretation: 
Consider $$P(x,y,z)p+Q(x,y,z)q=R(x,y,z)\tag1$$
and$$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}\tag2$$
Let $$z=\phi(x,y)\tag3$$
represents the solution of equation $(1)$. Then $(3)$ represents a surface whose normal at any point $~(x,~y,~z)~$ has direction ratios $~\frac{\partial z}{\partial x},~\frac{\partial z}{\partial y},~-1~$ i.e., $~p,~q,-1~$. 
Also we know that the simultaneous equation $(2)$ represents a family of curves such that the tangent at any point has direction ratios $~P,~Q,~R~$. 
Rewriting $(1)$ , we have $$P(x,y,z)p+Q((x,y,z)q+R(x,y,z)~(-1)=0\tag4$$
showing that the normal to surface $(3)$ at any point is perpendicular to the number of family of curves $(2)$ through that point. Hence the member must touch the surface at that point. Since this holds for each point on $(3)$, we conclude that the curves $(2)$ lie completely on the surface $(3)$ whose differential equation is $(1)$.
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Second Interpretation: 
We know that the curve whose equations are solutions of $$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}\tag{a}$$
are orthogonal to the system of the surfaces whose equation satisfies  $$P(x,y,z)~dx+Q(x,y,z)~dy+R(x,y,z)~dz=0~.\tag{b}$$
From the previous discussion we conclude that the curve of $(a)$ lie completely on the surface represented by $$P(x,y,z)p+Q((x,y,z)q=R(x,y,z)~.\tag{c}$$
Hence surfaces represented by $(b)$ and $(c)$ are orthogonal.

References:
$1.~~$ "Introductory Course in Differential Equations" by Daniel A. Murray
$2.~~$ "Ordinary and Partial Differential Equations" by M.D. Raisinghania
