Let x=a, y=2b, z=3c
1/a+1/b+1/c=1
WLOG assume that a≤b≤c
=>1/a≥1/b≥1/c
So, 3/a≥1
=>a≤3
As a≠1,
2 possible cases arise.
Case 1
a=2
1/b+1/c=1/2
Now, 2/b≥1/2
=>b≤4
We know, b≠1,2
Subcase 1
b=3
We get c=6
(a,b,c)=(2,3,6)
=>(x,y,z)=(2,6,18) -(i)
Subcase 2
b=4
We get c=4
So, (x,y,z)=(2,8,12) -(ii)
Case 2
a=3
So, 1/b+1/c=2/3
Now, 2/b≥2/3
=>b≤3
b≠1
Subcase 1
b=2
c=6
=>(x,y,z)=(3,4,18) -(iii)
Subcase 2
b=3
c=3
=>(x,y,z)=(3,6,18) -(iv)
So, the solutions are (2,6,18); (2,8,12); (3,4,18); (3,6,18)
PS- I know its quite long but its an easy approach