Location game between 3 firms
$(d)$ cannot be an answer because of the following.  If 1 and 2 are located as in $(d)$, then 3 wants to be located (i) either at $\frac{1}{3}-\varepsilon$ or (ii) at $\frac{2}{3}+\varepsilon$ because doing so yields almost $\frac{1}{3}$ while choosing $\frac{1}{2}$ yields $\frac{1}{2}\times\frac{1}{3}$.  So, 1 receives $\frac{1}{2}\times\frac{1}{3}$ if 3 chooses (i), or 2 receives $\frac{1}{2}\times\frac{1}{3}$ if 3 chooses (ii).  Then, 1 and 2 does not want to act according to $(d)$.

$(b)$ is the right answer. The reasoning is: if 1 and 2 are located as in the answer, 3 is indifferent of choosing anything in $\left[\frac{1}{4}, \frac{3}{4}\right]$ all of which yield $\frac{1}{4}$ and any other location yields a lower payoff.  In other words, each of 1 and 2 can guarantee himself at least $\frac{1}{4}$.   

Now, the question is: Does any of 1 and 2 want to deviate from $(b$)?  Suppose when both 1 and 2 played according to $(b)$, 3 chooses $\frac{1}{2}$ so that each of 1 and 2 obtains $\frac{3}{8}$.  Now, suppose that 1 deviates to a location $\frac{1}{4}-x$, where $x \in (0,1/4)$.  Then, he obtains $\frac{1}{4}-x + \frac{\frac{1}{2}-\left(\frac{1}{4}-x\right)}{2} =\frac{3}{8}-\frac{x}{2}$ which is clearly lower than $\frac{3}{8}$ as $x&gt;0$.  Thus, 1 does not want to deviate.  By a similar argument, 3 does not want to deviate too.  Hence, $(b)$ is an equilibrium