How to solve this system of Partial Differential Equations [5] $${\partial z \over \partial x}f(x,y,z)+{\partial z \over \partial y}g(x,y,z)=h(x,y,z)   $$
$${\partial z \over \partial x}F(x,y,z)+{\partial z \over \partial y}G(x,y,z)=H(x,y,z)   $$
$$f(x,y,z)={x-x_1 \over {\sqrt {(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} }}-{x-x_2 \over {\sqrt {(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} }} $$
$$g(x,y,z)={y-y_1 \over {\sqrt {(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} }}-{y-y_2 \over {\sqrt {(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} }} $$
$$h(x,y,z)={z-z_1 \over {\sqrt {(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} }}-{z-z_2 \over {\sqrt {(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} }}    $$
$$F(x,y,z)=(y-y_1)(z-z_2)-(y-y_2)(z-z_1)  $$
$$G(x,y,z)=(x-x_2)(z-z_1)-(x-x_1)(z-z_2) $$
$$ H(x,y,z)=(x-x_1)(y-y_2)-(x-x_2)(y-y_1) $$
$ x_1, y_1, z_1, x_2, y_2, z_2 $ Are constants
What is the standard approach to solving these equations? I've gone through my textbook, but the author has not discussed this case. Moreover, I'm not being able to word my problem correctly, as a consequence, the Google Searches on the Internet are resulting in utter failure. Please help
Thank You.
 A: This approach is not standard but this set of PDE can be solved using geometric arguments.
Let $\vec{p} = (x,y,z)$ be any generic point in $\mathbb{R}^3$.
Let $\vec{p}_i = (x_i,y_i,z_i)$, $\vec{r}_i = \vec{p} - \vec{p}_i$ and $r_i = |\vec{r}_i|$ for $i = 1,2$.
Given any function $f(x,y)$, the vector
$$\vec{N_f(x,y)} \stackrel{def}{=} \left(\frac{\partial f(x,y)}{\partial x}, \frac{\partial f(x,y)}{\partial y}, -1\right)$$
is proportional to the normal vector for the surface $z = f(x,y)$ at point $(x,y, f(x,y))$.
The set of given PDE can be rewritten as
$$\begin{align}
\vec{N}_f(x,y) \cdot \left[ \nabla (r_1 - r_2) \right]_{(x,y,f(x,y))} &= 0\\
\vec{N}_f(x,y) \cdot \left[ \vec{r}_1 \times \vec{r}_2 \right]_{(x,y,f(x,y))} &= 0
\end{align}
$$
Geometrically, solving the PDE is equivalent to finding a surface, in the form $z = f(x,y)$, such that for any point on it, the normal vector $N_f$ is perpendicular
to two vector fields:


*

*The vector field $\nabla (r_1 - r_2)$ which is proportional to the normal vectors on the families of two-sheet hyperboloid $r_1 - r_2 = const.$ which have foci at $\vec{p}_1$ and $\vec{p}_2$.

*The vector field $\vec{r}_1 \times \vec{r}_2$ which is proportional to the
normal vectors on the families of planes passing through the line joining $\vec{p}_1$ and $\vec{p}_2$.
It is well known on the plane, the families of ellipses and hyperbolas sharing the same foci are orthogonal to each other. Lift this phenomenon to $\mathbb{R}^3$, we immediately find 

The set of ellipsoid $r_1 + r_2 = const.$ with foci at $\vec{p}_1$ and $\vec{p}_2$ is a family of surfaces that "solves" the given PDE.

I'm lazy, I will leave the task to unwind the expression $r_1 + r_2 = const.$ to the explicit form $z = f(x,y)$ to you.
