= Solution
Take the grounded plane as $z=0$. The <method of images> replaces it, in $z>0$, by an <image charge> $-q$ at $z=-d$ and an <image charge> $+q$ at $z=-2d$. These make the <electric potential> zero on the plane; the <uniqueness theorem for the electrostatic potential> establishes the solution in the physical half-space.
At $z=3d/2$, both real <point charges> contribute upward <electric fields>, each of magnitude $kq/(d/2)^2$. The nearer <image charge> contributes downward and the farther one upward. Hence
$$
\mathbf E=\frac{kq}{d^2}\left(4+4-\frac4{25}+\frac4{49}\right)\widehat{\mathbf z}
=\boxed{\frac{9704}{1225}\frac{q}{4\pi\epsilon_0d^2}\widehat{\mathbf z}.}
$$
This is upward for $q>0$; retaining signed $q$ automatically handles the opposite sign. No self-<electric field> is involved because the evaluation point is distinct from every <point charge>.
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