I agree that finding the equations of the planes is a good idea. Remembering that a plane of the form has the same coefficients as its normal vector, we can get the normal vector by finding two vectors that lie on the plane and taking their cross product.
Two vectors that lie on the first plane are and . So the first normal vector is found by
So your first plane will have equation
Substituting in any of the points, such as (0, 0, 0), finds that and so the equation of your plane is , which as you already determined can be written as . So really this is a line in the x-y plane, that is then shifted vertically indefinitely in every direction. So as long as you have any two points which satisfy , then any value of z will be fine. So yes, this equation of your plane is possible and (0, 0, 1) lies on it.
I disagree with your calculation of the second plane. You have the points (1, 2, 0), (0, 0, 2) and (3, 0, 0) lying on it.
Two vectors which lie on the plane are and . So the normal vector is given by
So the equation of your plane will be , and substituting a point on the plane, like (0, 0, 2), gives . So the equation of your plane is
Now it would be a good idea to draw your region, so that you can decide which variables would be easiest to integrate over first.