Find the vector equation of the plane that is spanned by $\vec{v}=[-2, 1, \frac{1}{2}], \vec{w}=[4, -\frac{1}{5}, -1]$ and contains $[0, -1, 4]$ I was given the following problem:

Find the vector equation of the plane that is spanned by $\vec{v}=[-2, 1, \frac{1}{2}], \vec{w}=[4, -\frac{1}{5}, -1]$ and contains $[0, -1, 4]$


Note: Clearly $[a, b, ...]$ denotes a vector.

I'm really unsure about my procedure and wanted to see if it's correct. Here's what I attempted to do.

$I$. $\vec{n}:=[-2, 1, \frac{1}{2}] \times[4, -\frac{1}{5}, -1] = [-\frac{9}{10}, 0, -\frac{18}{5}]$ is a vector normal to $\vec{v}$ and $\vec{w}$. This means the cartesian equation of the plane that is spanned by $\vec{v}, \vec{w}$ is $-\frac{9}{10}x +0y-\frac{18}{5}z=c$. Because $(0, -1, 4)$ is a point in the plane we know
$$-\frac{9}{10}(0) -\frac{18}{5}(4)=c \iff c=-72$$
from which follows that the plane is defined by the equation
$$-\frac{9}{10}x +0y-\frac{18}{5}z=-72$$
or equivalently
$$-\frac{1}{10}x -\frac{2}{5}z=8$$
$II.$ Clearing for $z$ we have $z= 20-\frac{1}{4}x+0y$. From this we can build the vector equation of the plane as
$$\vec{g}(t)=[0, 0, 20]+t_1[1, 0, -\frac{1}{4}] + t_2[0, 1, 0]$$

I have no clue whatsoever about whether this is correct or not and could really use some validation/correction. The reasoning seems clear:
$i$. We find a cartesian equation for the plane, and ensure we are satisfying the condition that the plane passes through the point via finding the appropriate $c$.
$ii.$ We clear the cartesian equation for one of the variables ($z$ in this case) and build a vector equation that ensures the satisfaction of the "cleared" cartesian equation.
Does this make sense?

One more way:
Let the point [0,-1,4] be $\vec a$, then the eq. of the plain is $(\vec v \times \vec w).(\vec r-\vec a)=0$.Or one may also write
$(\vec v \times\vec w).\vec r=[\vec v, \vec w, \vec a].$
Edit:
Other (vector) eqn is
$(\vec r - \vec a)=p\vec v + q \vec w$, where $p,q$ are scalars.
