Assuming the algebraic rearrangement was correct, you go with

I:   x - 2y + 3z = 9
II: 0x +  y + 3z = 5
III: 2x - 5y + 5z = 17

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Then doing (-2)\*I = I' gives you

I': -2x + 4y - 6z = -18
II:  0x +  y + 3z =  5
III:  2x - 5y + 5z =  17

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As you can now see, you get a term of -2x in I' and of 2x in III. Thus if you add this I' to III, you get III' with x being eliminated:

I': -2x + 4y - 6z = -18
II:  0x +  y + 3z =  5
III':  0x -  y -  z = -1
But I'm mostly confused about where that -2 came from and what that second step is actually telling me to do
edit why did this post twice, i hope it doesn’t touch the other comment
The point of finding the row echelon form of a system of equations is doing certain operations that we are allowed to do (e. g. scalar multplication and adding/subtracting rows together) to rewrite the system into a form that we like.

The -2 is not something that followed from some formula or computation, but rather observation. Whoever solved the problem noticed that adding -2(row 1) to row 3, would get rid of the x term in the third row. Because we want the third row to not have a term with x on the left hand side, we decided that subtracting row 1 twice from the third row is a good idea because then the x from row 3 disappears.

So for a system like

3x-2y=2
4x+y=6

We could do something like multiply row 1 by -4 and multiply row 2 by 3 to get

-12x+8y=-8
12x+24y=18

Which after adding the first row to the second gives

-12x+8y=-8
0x+32y=10

Now divide the first row by - 12, and divide the second by 32 and you have succesfully rewritten the system into row echelon form
(arithmetic is not my strong suit so I might have messed up the example at some point. Hopefully you still get what I mean)