There are three test tubes with AlCl3, NaBr, KI solutions. How to use two reagents (excluding AgNO3) to determine what is in which test tube?

Decision:
1) Divide each solution in a test tube into 2 equal parts by pouring into another clean test tube;
2) Add alkali solution to the first 3 tubes with AlCl3, NaBr, KI. The reaction will occur only in a test tube with AlCl3 in the form of a precipitate:
AlCl3 + 3NaOH = 3KCl + Al (OH) 3 ↓
3) Add lead (II) nitrate to the second 2 test tubes (because we have already found a test tube with AlCl3 and we do not need it). The reaction will take place in both test tubes. Only one test tube will contain a white precipitate, which means there was NaBr in it; and in another test tube a yellow precipitate will fall out – it means that there was KI in it:
2NaBr + Pb (NO3) 2 = 2NaNO3 + PbBr2 ↓ (white)
2KJ + Pb (NO3) 2 = 2KNO3 + PbI2 ↓ (yellow).