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1. The condition is incorrect.
The correct condition must say that x, y and z are positive integers.
Otherwise the number of solution is more than infinite.
2. OK. Let us add it into the condition.
Then, if 360=2x×3y×5z, then xyz = 12, and it is much easier to solve.
The solutions are ("by the modulus of permutations", which doesn't matter, because the question is about the sum x+y+z invariant to permutations)
a) 1, 1, 12 ---> x + y + z = 1 + 1 + 12 = 14;
b) 1, 2, 6 ---> x + y + z = 1 + 2 + 6 = 9;
c) 1, 3, 4 ---> x + y + z = 1 + 3 + 4 = 8;
d) 2, 2, 3 ---> x + y + z = 2 + 2 + 3 = 7.
Answer. x + y + z = 14, 9, 8, 7.
If I missed something and you found it, then add it to the answer.
