If the 9-digit number 957x5y6z3 is divisible by 33, then what is the maximum value of (x+y+z)?

For a number to be divisible by 33, it must be divisible by both 3 and 11.
Divisibility by 3:
Sum of digits must be a multiple of 3.

Sum=9+5+7+x+5+y+6+z+3=35+x+y+z
For (35+x+y+z) to be divisible by 3, (x+y+z) can be 1,4,7,10,13,16,19,22,25.

Divisibility by 11:
Difference between sum of digits at odd places and even places must be 0 or a multiple of 11.

Sum of odd positions=9+7+5+6+3=30
Sum of even positions=5+x+y+z
Difference=30−(5+x+y+z)=25−(x+y+z)
For divisibility by 11:

25−(x+y+z)=0⟹x+y+z=25
or

25−(x+y+z)=11⟹x+y+z=14
Checking with the condition for divisibility by 3:
If x+y+z=25, then sum of digits = 35+25=60, which is divisible by 3.
Thus, the maximum value possible for (x+y+z) is 25.