= Solution
The <phase> step between adjacent slits is
$$
\delta=\frac{2\pi d\sin\theta}{\lambda}\simeq\frac\pi3.
$$
For equal <amplitudes> $a_0$, the six unmodified contributions form a closed <regular hexagon> in the <complex plane>:
$$
S_0=a_0\sum_{j=0}^{5}e^{ij\pi/3}=0.
$$
<Reversing one slit at an array minimum> replaces slit $j$'s contribution by its negative, so $S=S_0-2a_0e^{ij\pi/3}$. Consequently
$$
\boxed{I_{\rm screen}=|S|^2=4|a_0|^2=4I.}
$$
This is independent of which slit is reversed. The result uses the far-field, narrow-slit and <small-angle approximations>; retaining $\sin\theta$ changes the <phase> step only by about $7\times10^{-7}$ radians.
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