= Solution
Take the source at the origin and write $\rho^2=y^2+z^2$. Repeated reflection in $x=\pm1$ places the sources at $(2n,0,0)$ with strengths $(-1)^n$, $n\in\mathbb Z$. There are \b[countably infinitely many images]; $n=0$ is the original source. A convergent prescription groups adjacent opposite charges:
$$
\boxed{G(x,y,z;0)=-\frac1{4\pi}\sum_{m\in\mathbb Z}\left[\frac1{\sqrt{(x-4m)^2+\rho^2}}-\frac1{\sqrt{(x-4m-2)^2+\rho^2}}\right].}
$$
Each bracket is $O(|m|^{-2})$ on compact sets away from the source, so the paired series converges absolutely. The individual unsigned point-charge terms would not be absolutely summable. On $x=1$, reflection pairs the index $n$ with $1-n$; on $x=-1$, it pairs $n$ with $-1-n$. Their distances agree and signs oppose. Symmetric finite cutoffs leave only a vanishing far-tail contribution, proving both zero boundary values. All nonzero-index sources lie outside the slab. This is the centered-source version of the <Dirichlet Green function between parallel planes>, selected to decay at large transverse distance.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-2-image-source-arrangements.png]
{title=Image locations and strengths for an octant, an interior spherical source and two parallel plates}
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