Let x+1=a, y+2=b, and z+3=c be substituted. If a>2, then only if a=3, b=3, and c=4 are equal, and in this case y=z=1, it is contradictory by the problem condition. Thus a=2. If a=2, 1/b+1/c=5/12 is 1/b+2, c>3, so there is one greater than 5/24 in 1/b or 1/c, that is, b or c<5. The case work only works when x=1, y=2, and z=3. Q.E.D

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thdwlgh1229; Only; Not only the numbers (1,2,3), but also some numbers such as (2,1,1), (1,1,9) are also provided. We can show this as follows: Answer 4; (1,1,9), (1,2,3), (1,4,1), (2,1,1). It is obvious that x=<2, y=<4, z=<9, only (2,1,1) is satisfied in the case of x=2. In the case of x=1, we should see whether we can write 1/y+2 + 1/z+3 in the denominator of 12. Here y+2=3,4,6 may occur (due to 1=<y=<4) which means 3 solutions for x=1 and 1 solution for x=2. (This solution was written with google translate, there may be errors)

$a=x+1, b=y+2, c=z+3$, obviously, $2\leq a \leq 3$ By some casework, we have that $\boxed{(x,y,z)=\{(1,1,9),(1,2,3),(1,4,1),(2,1,1)\}}$

Let's substitute x + 1 = a, y + 2 = b, z + 3 = c. (1/a)+(1/b)+(1/c)=11/12. Let's consider the cases separately. case 1)a=1 :It is obviously impossible case 2)a=2 :(1/b)+(1/c)=11/12 - 1/2=5/12. (b,c)=(3,12),(12,3),(4,6),(6,4) ∴(x,y,z)=(1,1,9),(1,2,3),(1,4,1) case 3)a=3 :(1/b)+(1/c)=11/12 - 1/3=7/12. (b,c)=(2,12),(12,2),(3,4),(4,3) ∴(x,y,z)=(2,1,1) case 4)a>=4 :It does not satisfy the condition of the problem. (∵y and z become 0 or negative.) Since x, y, and z must be distinct, the only possible (x, y, z) is (1, 2, 3).