386
+5Z7
 XY3

The right column adds to 13 so there is 1 to carry to the
middle column, so we put a 1 to carry above it:

  1
 386
+5Z7
 XY3

Adding the second column:

Either there is or there isn't a 1 to carry to the leftmost column.

We consider the possibility that there is no carry to the leftmost column:

Then Z must be 0 and Y must be 9, and X must be 8:

  1
 386
+507
 893

Trouble is, 893 is not divisible by 3.  So this case is eliminated.

So there must be a carry of 1 to the leftmost column:

 11
 386
+5Z7
 XY3

Therefore X can only be 9

 11
 386
+5Z7
 9Y3

The sum XY3 which is 9Y3 must be divisible by 3.  Therefore
its sum of digits 9+Y+3 must be a multiple of 3.

Since the sum of the first and third digits of 9Y3 is 9+3=12,
which is a multiple of 3, Y must also be divisible by 3.
so Y = 0, 3, 6, or 9.

For Z to be as small as possible, Y must be as small as possible:
so Y must be 0.

 11
 386
+5Z7
 903

And so Z must be 1:

 11
 386
+517
 903

So the smallest possible value of Z is 1.

Edwin