= Solution
Let $\widehat{\mathbf e}_r$ point from the intersection to the real <point charge>. Reflection symmetry cancels the component of <electrostatic force> perpendicular to the wedge bisector. An <image charge> $q_j$ at angular separation $\Delta$ from the real <point charge> gives radial <force>
$$
F_{r,j}=kqq_j\frac{d(1-\cos\Delta)}{[2d^2(1-\cos\Delta)]^{3/2}}.
$$
The two adjacent negative <image charges> have $\Delta=\pm60^\circ$, distance $d$, and combined radial <force> $-kq^2/d^2$. The two positive <image charges> have $\Delta=\pm120^\circ$, distance $\sqrt3d$, and combined radial <force> $+kq^2/(\sqrt3d^2)$. The opposite negative <image charge> is at distance $2d$ and contributes $-kq^2/(4d^2)$. Thus
$$
\boxed{\mathbf F=-\frac{q^2}{4\pi\epsilon_0d^2}\left(\frac54-\frac1{\sqrt3}\right)\widehat{\mathbf e}_r.}
$$
The <electrostatic force> is towards the intersection, independently of the sign of $q$. One sums the five <image charge> <electric fields> and multiplies by $q$; the factor $1/2$ used in induced-charge <electrostatic energy> must not be inserted into this direct <force> calculation.
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