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.
Best Answer
Recall that if a plane $\pi$ in $\Bbb{R}^3$ has Cartesian equation $$ a x_1 + b x_2 + c x_3 = d $$ then the vector $ \mathbf{n} = \left(\begin{smallmatrix} a \\ b \\ c \end{smallmatrix}\right) $ is orthogonal to $\pi$. Also, recall that a parametric equation for $\pi$ has form $$ \mathbf{x} = \mathbf{x}_0 + t_1 \mathbf{v}_1 + t_2 \mathbf{v}_2 $$ where $\mathbf{x}_0$ is the vector of coordinates of any point on $\pi$ and $\mathbf{v}_1,\mathbf{v}_2$ are any two linearly independent vectors orhtogonal to $\mathbf{n}$.
Since you are already given $\mathbf{v}_1$ all you have to do is solve the system of linear equations $$ \begin{cases} 0 = \mathbf{n} \cdot \mathbf{v}_2 = a v_1^{(2)} + b v_2^{(2)} + c v_2^{(3)} \\ 0 = \mathbf{v}_1 \cdot \mathbf{v}_2 = v_1^{(1)} v_1^{(2)} + v_2^{(1)} v_2^{(2)} + v_3^{(1)} v_3^{(2)} \\ d = \mathbf{n} \cdot \mathbf{x}_0 = a x_1^{(0)} + b x_2^{(0)} + c x_3^{(0)} \end{cases} $$