= Solution
Let $\overline{\mathbf r'}=(x',y',-z')$ be reflection of the interior source across the equatorial plane. Using the ball <Green's function> from part (c), form
$$
\boxed{G_H(\mathbf r,\mathbf r')=G_B(\mathbf r,\mathbf r')-G_B(\mathbf r,\overline{\mathbf r'})}.
$$
Both terms vanish on the unit sphere. On the flat disk $z=0$, distances to reflected points coincide, and the same is true of their inversion images, so the difference is zero there as well. Only $\mathbf r'$ is inside the upper half-ball; the reflected source is in the lower half-ball, and both inversion images lie outside the full ball. Thus there is exactly one interior delta source, of the required strength.
Explicitly, with $s=|\mathbf r'|$ and stars denoting inversion in the unit sphere,
$$
\boxed{G_H=-\frac1{4\pi}\left[
\frac1{|\mathbf r-\mathbf r'|}-\frac1{s|\mathbf r-\mathbf r'^*|}
-\frac1{|\mathbf r-\overline{\mathbf r'}|}+\frac1{s|\mathbf r-\overline{\mathbf r'}^*|}
\right]}.
$$
This is the <Dirichlet Green function inside a hemisphere>, constructed from four image-source terms.
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