Let T be the solid formed by taking all nonnegative (x,y,z) under the plane 3(x-2) + 2y + 6z = 0. It is the tetrahedron with vertices (0,0,0), (2,0,0), (0,3,0), and (0,0,1).

To determine the limits of integration, we take the outermost variable and work inward. The integral will have the general form

- Determine the maximum and minimum values of the outermost variable. These will be the limits of integration on the first integral sign.
Since x is between 0 and 2, the limits of the first integral sign are 0 and 2.

- View a slice formed by keeping the outermost variable constant. Now determine the maximum and minimum values of the middle variable within that slice in terms of the outermost variable. This will give the limits of integration for the middle integral. Note that if the maximum and minimum values depend on where the slice is taken, you will need to split the integral.
The slice formed by keeping x constant is a triangle.

The line formed where the plane 3(x-2) + 2y + 6z = 0 intersects the xy plane is y = -3x/2 + 3. This is found by letting z = 0 in the equation 3(x-2) + 2y + 6z = 0.

Therefore we can label the bottom vertex of the triangle formed by the slice.

Since y is between 0 and -3x/2 + 3, the limits of the middle integral sign are 0 and -3x/2 + 3.

- Finally, using the same slice, determine the range of the innermost variable in terms of the other two variables. This will give the limits of integration for the inner integral. Note that if the range of the innermost variable changes within the slice, you will need to split the integral.
The hypotenuse of the triangle formed by taking a slice of the solid with x constant lies on the plane 3(x-2) + 2y + 6z = 0. We solve this equation for z and get z = -x/2 - y/3 + 1.

Since z is between 0 and -x/2 - y/3 + 1, these are the limits of the innermost integral sign.

Thus the final integral is as follows:

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