$dS$ is a surface element, a differential sized part of the surface $S$.
It is usually oriented, positive if its normal $n$ is outward pointing (e.g. if $S$ is the boundary of a volume).
$$
dS = n \lVert dS \rVert
$$
I have seen both
$$
d\mathbf{S} = \mathbf{\hat N}dS = \pm
(\mathbf{\frac {n}{|n|}})(\mathbf{|n|}) dudv
$$
(for parametric surfaces), and
$$
d\mathbf{S} = \mathbf{\hat N}dS = \pm \frac{\nabla
G(x,y,z)}{G3(x,y,z)}dxdy
$$
for level surfaces.
For those examples $\lVert dS \rVert = du \, dv$ and $\lVert dS \rVert = dx \, dy$. The other parts are the more or less complicated normal vectors of those surface elements.
$$ dS = \frac{dxdy}{|\mathbf{\hat N}\bullet\mathbf{j}|} =
\frac{\sqrt{14}}{2} dxdz $$
The integration is along the $x$-$z$ plane, while the surface,
$$
S: x+2y+3z = 6 \quad n = (1,2,3)^t/\sqrt{14}
$$
which is a plane as well, is not parallel to the $x$-$z$ plane.
The area of the projection $P_y S$ has to be adjusted, to give the correct area $\lVert S \rVert$ for $S$. We want
$$
\lVert S \rVert
= \int\limits_S \lVert dS \rVert
= \int\limits_S \lVert n\,du\,dv \rVert
= f \lVert P_y S \rVert
= f \int\limits_{P_y S} \lVert dx \, dz \rVert
$$
In your example they simply take $f = 1/\lVert n \cdot e_y\rVert$.
Let us check this: First we look for unit vectors $u$ and $v$ orthogonal to $n$ and each other.
$$
0 = n \cdot a = (1, 2, 3)^t / \sqrt{14} \cdot (2, -1, 0)^t \quad
e_u = (2, -1, 0)^t / \sqrt{5} \\
e_v = n \times e_u = (3, 6, -5)^t / \sqrt{70}
$$
These are unit vectors, so the area of the square between $e_u$ and $e_v$ is 1.
Now these unit vectors have the projections on the $x$-$z$ plane:
$$
u_p = P_y e_u = (2, 0, 0)^t/\sqrt{5} \quad
\lVert u_p \rVert = 2/\sqrt{5} \\
v_p = P_y e_v = (3, 0, -5)^t/\sqrt{70} \quad
\lVert v_p \rVert = \sqrt{34/70} = \sqrt{17/35} \\
$$
where $P_y a = a - (a\cdot e_y) e_y$ for a vector $a$. The area of the projection is
$$
\lVert u_p \times v_p \rVert
=
\lVert ((2, 0, 0)^t/\sqrt{5}) \times ((3, 0, -5)^t/\sqrt{70}) \rVert
=
\lVert (0, 10, 0)^t/\sqrt{350} \rVert = 2 /\sqrt{14}
$$
This should explain the factor $\sqrt{14}/2$.
What is missing is a derivation for the shorter
$$
\lVert P_y u \times P_y v \rVert =
\lVert n \cdot e_y \rVert \, \lVert u \rVert \, \lVert v \rVert
$$
The standard way is to parametrise the surface. The surface can be parametrised in the following way: $\mathbf{r}(u,v)=\langle u, 4-2u, v \rangle$, where $u\in (0,2)$ and $v\in(0,4)$. Then, take partial derivatives $\mathbf{r}_u=\langle 1,-2,0\rangle$ and $\mathbf{r}_v=\langle 0, 0, 1\rangle$; consequently, $\mathbf{r}_u\times\mathbf{r}_v=\langle -2, -1, 0\rangle$. This will give you one choice of normal vector $\mathbf{n}$, but whether this is the correct orientation, it is dependent on the question(the other being $-\mathbf{n}$). That is why I ask for the correct orientation.
We compute the surface integral in the following way: $\int_S\mathbf{F} \mathrm{d}S=\iint_D\mathbf{F}\cdot(\mathbf{r}_u\times\mathbf{r}_v)\mathrm{d}A$, where $D$ is comprised of all the $(u,v)$ in domain. Here, the equation gives $\iint_D\langle 4-2u,2u,-v\rangle\cdot\langle -2, -1, 0\rangle\mathrm{d}A=\int_0^4\int_0^2(-8+2u)\mathrm{d}u\mathrm{d}v=-48$
Best Answer
It works well too. Try better with $z-\sqrt{x^2+y^2}=0$
$\vec n=(-z_x,-z_y,1)=\left(\dfrac{-x}{\sqrt{x^2+y^2}},\dfrac{-y}{\sqrt{x^2+y^2}},1\right)=(\cos\theta,\sin\theta,1)$
$d\mathbf S=\vec ndS=(-z_x,-z_y,1)dxdy=(-\cos\theta,-\sin\theta,1)rdrd\theta $
$=(-r\cos\theta,-r\sin\theta,r)drd\theta$