Use the known hydrolysis patterns of interhalogen compounds

In interhalogen compounds, the more electronegative halogen usually forms the hydrogen halide on hydrolysis, while the central/less electronegative atom forms an oxyacid or oxyanion.

Hydrolysis of $ICl$

The hydrolysis reaction is: $ICl + H_2O \rightarrow HOI + HCl$

The oxy-species formed is hypoiodous acid, whose anion is: $HOI \leftrightarrow H^+ + IO^-$

So, complete hydrolysis of $ICl$ gives $IO^-$.

Hydrolysis of $ClF_3$

Complete hydrolysis of chlorine trifluoride is: $ClF_3 + 2H_2O \rightarrow HClO_2 + 3HF$

The oxyacid formed is chlorous acid, whose anion is: $HClO_2 \leftrightarrow H^+ + ClO_2^-$

So, complete hydrolysis of $ClF_3$ gives $ClO_2^-$.

Hydrolysis of $BrF_5$

Complete hydrolysis of bromine pentafluoride is: $BrF_5 + 3H_2O \rightarrow HBrO_3 + 5HF$

The oxyacid formed is bromic acid, whose anion is: $HBrO_3 \leftrightarrow H^+ + BrO_3^-$

So, complete hydrolysis of $BrF_5$ gives $BrO_3^-$.

Match with options

Thus, respectively, the products are: $IO^-,\quad ClO_2^-,\quad BrO_3^-$

This matches Option A.

Compare with stored correct answer

Stored correct answer: A
Derived answer: A

Hence, they agree.