Solving system of three partial differential equations I have three differential equations:
$$\dfrac{\partial H}{\partial y}-\dfrac{\partial G}{\partial z}=A$$
$$\dfrac{\partial F}{\partial z}-\dfrac{\partial H}{\partial x}=B$$
$$\dfrac{\partial G}{\partial x}-\dfrac{\partial F}{\partial y}=C$$
where $A$, $B$ and $C$ are functions of $(x,y,z)$
How shall I write $F,G,H$ in terms of $A, B, C$?

EDIT: There is the condition : $\nabla.(\nabla \times \vec{A})=0$
Now how shall I proceed to get a solution?

 A: You want to solve
$$ \nabla \times \textbf{V} = \textbf{U} $$
with the condition $$ \nabla \cdot \textbf{U} = \frac{\partial A}{\partial x} + \frac{\partial B}{\partial y} + \frac{\partial C}{\partial z} =0 $$
Note that since the curl of any gradient field is $0$, the solution will not be unique, i.e. if $\textbf{V}$ is a solution, then another solution is $\bar{\textbf{V}} = \textbf{V} + \nabla \phi$, where $\phi(x,y,z)$ is some scalar function.
It's enough to find one particular solution, then add on an arbitrary gradient field at the end. A solution can be found by setting one of the components of $\textbf{V}$ to $0$. Let's say $H=0$ so the system is reduced to
\begin{align} 
- \frac{\partial G}{\partial z} &= A \\
\frac{\partial F}{\partial z} &= B \\
\frac{\partial G}{\partial x} - \frac{\partial F}{\partial y} &= C
\end{align}
Then integrate the first two equations
\begin{align} G &= - \int A\ dz + g(x,y) \\ F &= \int B dz + f(x,y) \end{align}
At this point, the two arbitrary functions leftover from integration should be set so that the third equation is satisfied
$$ \frac{\partial G}{\partial x} - \frac{\partial F}{\partial y} = -\int \left(\frac{\partial A}{\partial x} + \frac{\partial B}{\partial y} \right)dz + \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} = C $$
Suppose that
$$ -\int \left(\frac{\partial A}{\partial x} + \frac{\partial B}{\partial y} \right) dz = C + h(x,y) $$
where $h(x,y)$ is a known function, then we must have
$$ \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} + h = 0 $$
To make it easy again, we can arbitrarily set $g=0$ and $f = \int h\ dy$
