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From schaum's vector analysis:

I project the differential area $dS$ of the plane onto the $xy$ plane, then

$dxdy = dS (|hat n. hat k|)$ Where $hat n$ is the normal vector to $dS$

then $dS = frac{dxdy}{ |hat n. hat k| }$

$ hat n = frac{ nabla S}{ | nabla S | } = frac{2}{3} hat i + frac {1}{3} hat j + frac{2}{3} hat k$

$nablatimesvec F = 3 hat i - hat j - 2 hat k$

$(nablatimesvec F) . hat n = frac{1}{3} $

$dS = frac{3}{2} dxdy$

Then, $iint_S nablatimesvec F cdot hat n $ $dS$ = $ frac{1}{2} iint_S dxdy = frac{1}{2} iint_0^2 dxdy$ = $ frac{1}{2} int_0^1 2 dx = 1$

Now this is is when the boundaries are $x=0, x=1, y=0, y=2$

But when the boundaries are $x=0 , y=0, z =0$ , as follows:

It's all the same steps except for:

$iint_S nablatimesvec F cdot hat n = frac{1}{2} iint_S dxdy = frac{1}{2} iint_0^{6-2x} dxdy = frac {1}{2} int_0^3 6-2x dx = frac {9}{2}$

Now what I don't understand is:
In the first part, we treated $y$ as changing independently of $x$ from $y=0$ to $y=2$

In the second part, we treated $y$ as dependent on $x$ by the function $y=6-2x$ and integrated from $y=0$ to $y=6-2x$

Why is that? When do we substitute $y$ in as a function of $x$ and integrate like in the second part, and when not to? The problem is I cannot visualize 3 planes intersecting each other and visualize $S$ , so how can I understand it?

In the first problem, you are integrating the surface of the plane $mathcal{P}: 2x+y+2z=6$ over the rectangle $0leq xleq 1, 0leq yleq 2$. To parametrize the rectangle, simply use $x,y$ as parameters and integrate as you did.

For the second problem, you are integrating the surface of $mathcal{P}$ above the $xy-$plane. Imagine the plane $mathcal{P}$ as a sheet of paper slicing into the $xy-$plane. The sheet creates a shadow over the $xy-$plane, which will be in the shape of a triangle. To see this, let $z=0$ to see what the plane looks like in the $xy-$plane. We'll get
$$2x+y=6$$
Now, this is the line $y=6-2x$ which, when bounded by the lines $x=0$ and $y=0$, gives us a triangle. The triangle is a simple region and can be parametrized, for example, by
$$0leq xleq 3$$
$$0leq yleq 6-2x$$
which explains the solution to the second problem.

To address your point about $y$ being "independent" from $x$ in the first integral and not the second, it's because the first region is a rectangle, which can be parametrized by all constant bounds (this is the easiest case). However, for more general regions (like the triangle in problem $2$), you cannot express them as
$$aleq xleq b$$
$$cleq y leq d$$
but for example
$$aleq xleq b$$
$$f(x)leq yleq g(x)$$
where $y$ is bounded by two continuous functions of $x$. In this problem, those two functions are $f(x)=0$ and $g(x)=6-2x$.

Drawing boundaries on a graph helps a lot to understand the limits of the integrand.
For second part Q67), after drawing the boundaries on a xy plane, we get intuition for the boundaries pretty well, https://www.desmos.com/calculator/19sfc9hohv ,here vertical strip
changes from y=0 to y=-2x+6 and that strip stretches within x=0 to x=3.

But for first part Q66), after drawing the boundaries on xy plane, we get a rectangular region, where y(plane in 3D but a straight line in 2D) is obviously(intuition from graph) independent of x. Vertical strip changes from y=0 to y=2 and stretch that within x=0 to x=1 graph at https://www.desmos.com/calculator/xnho1f9uqu