If all of x, y, and z are 7 or bigger then we have 1/x+1/y+1/z<=1/7+1/7+1/7<7/15.  So at least one of x, y, or z is 6 or less.

Without loss of generality assume x<=y<=z.  Then x is one of 3, 4, 5, or 6.

Case 1: x=3.  Then 1/y+1/z=2/15

I can grind the algebra a bit to rewrite this equation into (2y-15)*(2z-15)=15^2=225.

The factorizations of 225 are (1,225), (3,75), (5,45), (9,25) and (15,15).  Then y and z are the corresponding pairs (8,120), (9,45), (10,30), (12,20), and (15,15).

The first five solutions with x<y<z are (3,8,120), (3,9,45), (3,10,30), (3,12,20), and (3,15,15).

Case 2: x=4.  Then 1/y+1/z=13/60

I can grind the algebra a bit to rewrite this equation into (13y-60)*(13z-60)=60^2=3600.  There are a lot of possible pairs but we need factorizations with each factor congruent to 5 mod 13 to have integer solutions.

The valid factorizations of 3600 are (5,720) and (18,200).

Then y and z are the corresponding pairs (5,60) and (6,20).

The next two solutions with x<y<z are (4,5,60) and (4,6,20).

Case 3: x=5.  Then 1/y+1/z=4/15

I can grind the algebra a bit to rewrite this equation into (4y-15)*(4z-15)=15^2=225.  This time we need factorizations congruent to 1 mod 4.

The valid factorizations of 225 are (1,225), (5,45), and (9,25).

Then y and z are the corresponding pairs (4,60), (5,15), and (6,10).

The first pair has x>y and is a repeat from case 2.

Case 4: x=6.  Then 1/y+1/z=3/10

I can grind the algebra a bit to rewrite this equation into (3y-10)*(3z-10)=10^2=100.  This time we need factorizations congruent to 2 mod 3.

The valid factorizations of 100 are (2,50) and (5,20).

Then y and z are the corresponding pairs (4,20) and (5,10).

Both of these pairs have x>y and are repeats from earlier cases.

No new solutions in this case.

In summary, the solutions of positive integers with x<=y<=z are (3,8,120), (3,9,45), (3,10,30), (3,12,20), (3,15,15), (4,5,60), (4,6,20), (5,5,15) and (5,6,10).