value of x,y,z

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

@Sambit your solution is wrong since a,b, and c need not be integers. I am showing a counter example:
\frac{1}{6}+\frac{2}{3}+\frac{1}{6}=1 Here x=6 y=3 and z=18
but a=6 b=3/2 and z=6