How to find the limits of $y$ in this surface integral?

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Integrate $G(x,y,z) = xyz$ over the triangular surface bounded by the point

$(1,0,0), (0,2,0)$ and $(0,1,1)$

Now I can calculate the equation of plane formed by the three points is :

$2x + y + z = 2$ but I am little confused on how should I calculate the limits for $x$ and $y$

Suppose I take the projection in $x-y$ plane then $z = 0$ then I get

$2x + y = 2$ So $x$ varies from $0$ to $1$ while $y$ varies from $0$ to $2-2x$

However in my book The limits of $y$ are given from $1-x$ to $2-2x$. I do not understand where this $1-x$ comes from ?

Can anyone please help me?

Thank you.

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In the $xy$-plane, $y$ is bounded by the two lines. One of them is $y=2-2x$ and the other is the line passing through (1,0,0) and (0,1,0), which is the $xy$-projection of the side passing through (1,0,0) and (0,1,1).

Thus, the other limit is given by,

$$y = 1-x$$

i.e. the line passing through (1,0,0) and (0,1,0).