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.