This is a variation on http://mathoverflow.net/questions/180360 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.

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*Example:* $A$ and $C$ are two infinite parallel planes, $A$ is at $z=-a$ and $C$ is at $z=c$, with $a+c=d$ the separation of the planes. The particle starts at $z=0$. The potential in this case can be obtained by the method of image charges, as an infinite alternating series. Summing this series to obtain $P_{A|C}$ is a bit tricky, a reliable way to do this is described by Marcus Zahn in <A HREF="http://www.rle.mit.edu/cehv/documents/16-AmericanJournalofPhysics441132-11341976.pdf">Point charge between two parallel grounded planes.</A> (The result we need from that paper is the total induced charge on each plane.) In this way one obtains 
$$P_{A|C}=\frac{c}{a+c}.$$