= Solution
Write $s=|\mathbf r'|$ and $\mathbf r''=a^2\mathbf r'/s^2$. If $|\mathbf r|=a$, expansion of squared <Euclidean distances> gives
$$
|\mathbf r-\mathbf r''|^2=\frac{a^2}{s^2}|\mathbf r-\mathbf r'|^2.
$$
Consequently $|\mathbf r-\mathbf r''|=(a/s)|\mathbf r-\mathbf r'|$, and the two terms in $H$ cancel. This <method of images> gives the <Dirichlet Green function inside a sphere>:
$$
\boxed{H(\mathbf r,\mathbf r')=-\frac1{4\pi}\left[\frac1{|\mathbf r-\mathbf r'|}-\frac{a/s}{|\mathbf r-\mathbf r''|}\right].}
$$
The image point is outside the sphere, so only the physical source contributes a <Dirac delta distribution> inside. As $s\to0$, the image position tends to infinity and its coefficient also diverges, but their ratio has the finite limit
$$
\frac{a/s}{|\mathbf r-\mathbf r''|}\longrightarrow\frac1a.
$$
Thus
$$
\boxed{H(\mathbf r,0)=-\frac1{4\pi|\mathbf r|}+\frac1{4\pi a}.}
$$
The apparent singularity of the image construction at a central source is removable; the physical source singularity at $\mathbf r=0$ remains.
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