Step 1: What the test must achieve.
We are separating three classes of amine at once, so a useful reagent has to behave in three clearly different ways. Only one of the listed reagents manages that, and it is benzenesulphonyl chloride, the Hinsberg reagent.
Step 2: How it behaves with each amine.
A primary amine reacts and the product still keeps an N to H bond, so that product dissolves when KOH is added. A secondary amine reacts too, but its product has no N to H bond left, so it stays undissolved in KOH. A tertiary amine has no N to H bond to begin with, so it simply does not react. Three amines, three different outcomes.
Step 3: Why the other choices fail.
The carbylamine reagent ($\mathrm{CHCl_3}$ with alcoholic KOH) only flags primary amines, and benzoyl chloride reacts with both primary and secondary, so neither one tells all three apart.
Step 4: Conclusion.
The single reagent that distinguishes all three is $\mathrm{C_6H_5SO_2Cl}$.
\[ \boxed{\text{Option (A): } \mathrm{C_6H_5SO_2Cl}} \]