Effects Guiding Electrophilic Aromatic Substitution
The first thing I notice is the hydroxyl group, a very powerful activating group and an ortho-para director.  The hydroxyl group is such a strong activator that if you take a solution of phenol and drip bromine into it, the phenol will react with the bromine so rapidly that the bromine color will be immediately removed from the solution as 2,4,6-tribromophenol is formed.

The next thing I notice is the very bulky t-butyl group, also an activator, but much milder than hydroxyl and also a mild ortho-para director.

Next I notice that the two substituents are arranged meta to one another.  There is one position in between the two groups that will be sterically blocked (double-ortho effect); the position on the other side of the t-butyl group will also be sterically blocked, but not as bad.  Also the directing effect of the hydroxyl group will overwhelm the directing effect of the t-butyl group.

Finally, in the halogen series size increases from chlorine to iodine.

In the reaction with iodine, the position ortho to the hydroxyl group and para to the t-butyl group is the only position that is electronically activated and sterically accessible to the bulky iodine.

In the reaction with smaller bromine, the position ortho to the hydroxyl and para to the t-butyl and the position para to the hydroxyl and ortho to the t-butyl are electronically activated by both groups and sterically accessible.  The "double-ortho" position is still sterically blocked.

With the smaller chlorine atom, all 3 positions - even the "double ortho" - position are sterically accessible to the smaller chlorine.

Based on this reasoning, I would select answers A, B, and C.  I would probably also include D because the results are (weakly) reinforced by the electronic effect of the t-butyl group.