= Solution
Define source strength $m$ here as the volume <flux> through a small sphere around the source. For incompressible <potential flow>, $\mathbf u=\nabla\phi$ and the governing conditions are
$$
\Delta\phi=m\delta(\mathbf x-\mathbf x_s),\qquad
\partial_z\phi=0\text{ on }z=0,\qquad \phi\to0\text{ at infinity}.
$$
Equivalently the <Laplace equation> holds away from the source, with local behavior $\phi\sim-m/(4\pi|\mathbf x-\mathbf x_s|)$. The boundary condition expresses impermeability. The <method of images> uses an equal source below the plane:
$$
\boxed{\phi(\mathbf x)=-\frac m{4\pi}\left(\frac1{\sqrt{x^2+y^2+(z-a)^2}}+\frac1{\sqrt{x^2+y^2+(z+a)^2}}\right).}
$$
The image singularity lies outside the fluid, the two vertical <derivatives> cancel on the wall, and the source has the specified outward <flux>.
Let $s=\sqrt{x^2+y^2}$. On the wall $u_z=0$ and the tangential radial velocity is $u_s=ms/[2\pi(s^2+a^2)^{3/2}]$. With ambient <pressure> $p_\infty$, the steady <Bernoulli equation> in this <irrotational flow> gives
$$
\boxed{p(s,0)=p_\infty-\frac{\rho m^2s^2}{8\pi^2(s^2+a^2)^3}.}
$$
The fluid <pressure> is reduced everywhere on the wall except at its axis and in the infinite-distance limit. The finite hydrodynamic force is the excess over the uniform ambient-pressure force. Integrating its magnitude over the plane yields
$$
F=2\pi\int_0^\infty[p_\infty-p(s,0)]s\,ds
=\frac{\rho m^2}{4\pi}\int_0^\infty\frac{s^3\,ds}{(s^2+a^2)^3}.
$$
Putting $u=s^2$ gives the last <integral> as $\tfrac12\int_0^\infty u(u+a^2)^{-3}du=1/(4a^2)$. Hence
$$
\boxed{F=\frac{\rho m^2}{16\pi a^2},\qquad \mathbf F=F\mathbf e_z.}
$$
Relative to ambient <pressure> on the other side, the boundary is \b[attracted toward the source]. This is the <hydrodynamic attraction of a plane wall to a point source>.
There is an alternative source-strength convention: if $m$ denotes the coefficient in $u_r=m/r^2$ rather than total volume <flux>, replace the <flux> above by $4\pi m$. Then $\phi=-m(1/r_++1/r_-)$, $p-p_\infty=-2\rho m^2s^2/(s^2+a^2)^3$, and $F=\pi\rho m^2/a^2$. The physical result is identical once the normalization is fixed.
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