The smallest possible total is 356 = 107 + 249, and the largest sum possible is 981 = 235 + 746, or 657 + 324. 
The middle sum may be either 720 =134+586, or 702 = 134 + 568, or 407 = 138 + 269. 
The total in this case must be made up of three of the figures 0, 2, 4, 7, but no sum other than the three given can possibly be obtained. 
We have therefore no choice in the case of the first locker, an alternative in the case of the third, and any one of three arrangements in the case of the middle locker. 

Here is one solution:

Of course, in each case figures in the first two lines may be exchanged vertically without altering the total, and as a result there are just 3,072 different ways in which the figures might be actually placed on the locker doors. 
I must content myself with showing one little principle involved in this puzzle. 
The sum of the digits in the total is always governed by the digit omitted. ⁹/₉ - ⁷/₁₀ - ⁵/₁₁ - ³/₁₂ - ¹/₁₃ - ⁸/₁₄ - ⁶/₁₅ - ⁴/₁₆ - ²/₁₇ - ⁰/₁₈. Whichever digit shown here in the upper line we omit, the sum of the digits in the total will be found beneath it. 
Thus in the case of locker A we omitted 8, and the figures in the total sum up to 14. 
If, therefore, we wanted to get 356, we may know at once to a certainty that it can only be obtained (if at all) by dropping the 8.