Think about it in terms of vectors. The vector equation of the plane is
$$ (x,y,z) = (1,4,8) + r(2,5,8) + s(3,6,9). $$
In other words, the plane passes through $(1,4,8)$, and is spanned by $(2,5,8)$ and $(3,6,9)$. Therefore,
$$ (x,y,z)-(1,4,8) = (x-1,y-4,z-8) = r(2,5,8) + s(3,6,9), $$
so it is a linear combination of $(2,5,8)$ and $(3,6,9)$. Or if you prefer, the parallelepiped with edges given by $(x-1,y-4,z-8)$, $(2,5,8)$ and $(3,6,9)$ has zero volume. Either one of these means that the determinant
$$\begin{vmatrix}2&3&x-1\\5&6&y-4\\8&9&z-8\end{vmatrix}$$
vanishes (the vectors don't form a basis of $\mathbb{R}^3$ if and only if they are all in the same plane, so this also means that you can't always solve
$$ a(x-1,y-4,z-8) + b(2,5,8) + c(3,6,9) = (e,f,g), $$
or
$$ \begin{pmatrix}2&3&x-1\\5&6&y-4\\8&9&z-8\end{pmatrix} (b,c,a) = (e,f,g), $$
so the matrix on the left must be singular.
Hopefully at least one of those explanations makes sense to you.
The idea is that the determinant is alternating multilinear: what this actually means is that if one column is made of linear combinations of the other columns, the determinant must be zero.
More explicitly, think of it as a map $\det$ that takes $n$ vectors in $\mathbb{R}^n$ and spits out a real number. It is also linear in each variable (vector), and the sign of this real number is changed if two of the vectors swap places: e.g. in $\mathbb{R}^3$,
$$ \det{(\mathbf{a},\mathbf{b},\mathbf{c})} = -\det{(\mathbf{b},\mathbf{a},\mathbf{c})} $$
and so on. This means firstly that
$$ \det{(\mathbf{a},\mathbf{a},\mathbf{c})} = -\det{(\mathbf{a},\mathbf{a},\mathbf{c})} $$
(swapping $\mathbf{a}$ with $\mathbf{a}$), and so
$$ \det{(\mathbf{a},\mathbf{a},\mathbf{c})} = 0 $$
for any $\mathbf{a}$ and $\mathbf{c}$. Secondly, the linearity gives
$$ \det{(r\mathbf{b}+s\mathbf{c},\mathbf{b},\mathbf{c})} = r\det{(\mathbf{b},\mathbf{b},\mathbf{c})}+s\det{(\mathbf{c},\mathbf{b},\mathbf{c})} = 0, $$
which is why if one column is a linear combination of the others, the determinant is zero.
If you have a pair of skew lines with direction vectors ${\bf a}$ and ${\bf b}$, then since they are skew, their direction vectors are not parallel. Non-parallel vectors will always yield a nonzero cross product. So ${\bf n} = {\bf a} \times {\bf b}$ will (for skew lines) always be a nonzero vector.
So just as with any nonzero vector, you can use ${\bf n}$ as a normal for a plane. But to determine a plane you need a normal vector and a point.
If you choose a point on one of the skew lines, you'll get a plane containing that line and the other skew line will be parallel to your plane.
If you choose a point not on either of the skew lines, you'll get a plane which is parallel to both of the lines but contains neither line.
Best Answer
The span of $(1,1,0)$ and $(0,2,0)$ consists of the $x$-$y$-plane, or in other words: all vectors with $z = 0$. So your equation is fine.
You can test if a linear combination $u$ is part of the plane, by checking if it fulfills the normal equation of the plane $n \cdot u = 0$: $$ n \cdot (s (1,1,0) + t (0,2,0)) = s ((0,0,2) \cdot (1,1,0)) + t ((0,0,2) \cdot (0,2,0)) = 0 \quad (s, t \in \mathbb{R}) $$