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.

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 Point charge between two parallel grounded planes. (The result we need from that paper is the total induced charge on each plane.) In this way one obtains $$P_{A|C}=\frac{c}{a+c}.$$

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.

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 Point charge between two parallel grounded planes. (The result we need from that paper is the total induced charge on each plane.) In this way one obtains $$P_{A|C}=\frac{c}{a+c}.$$

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 $P_C$.

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.

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 Point charge between two parallel grounded planes. (The result we need from that paper is the total induced charge on each plane.) In this way one obtains $$P_{A|C}=\frac{c}{a+c}.$$