This is a variation on Brownian motion and hitting a Quadrilateral and can be reduced in a similar way to the solution of a problem in electrostatics. Let me assume that the Brownian motion starts at the origin and ask for the probability $P_{A|C}$ that the particle eventually hits the boundary $A$ without first hitting the boundary $C$.
In the equivalent electrostatic problem the boundaries $A$ and $C$ are grounded (potential $\Phi=0$) and the probability $P_{A|C}$ is obtained by integrating (minus) the electric field $\nabla\Phi$ over the boundary of $A$:
$$\nabla^2\Phi(\vec{r})=-\delta(\vec{r}),\;\;\Phi(\vec{r})=0\;\;\text{for}\;\;\vec{r}\in A,C$$
$$P_{A|C}=\int_A \frac{\partial\Phi}{\partial\vec{r}}\cdot\hat{n}\;dS,$$
with $\hat{n}$ a unit vector normal to $A$ and pointing outward.
I hope this at least answers Ilya's question on "the couple of PDE's that the solution satisfies". No hope for a simple closed-form solution for arbitrary objects, I'm afraid. The electrostatic analogue also gives no hope to express the solution with two grounded planes in terms of the solution with a single grounded plane, so I don't think $P_{A|C}$ can be expressed in terms of the individual hitting probabilities $P_A$ and $P_C$.