$\text{Remark:}$ I notate vectors as $\langle a_1, a_2,\ldots\rangle$
I was given the following problem.
Find a vector equation for the plane $S$ spanned by $\langle -1, 0, 4\rangle, \langle 2, 3, 10\rangle$ and containing the point $P_0 = (2, 3, -5)$.
I am new to multivariate calculus and I self study. The lack of teachers forces me to post my solutions here in seek of correction/validation and suggestions. I will show what I did and explain a particular aspect of the problem that troubled me.
$I$. Let $\vec{v} = \langle -1, 0, 4\rangle, \vec{w} = \langle 2, 3, 10\rangle$. Then the normal vector of the plane spanned by $\vec{v}, \vec{w}$ is
\begin{align} \vec{n} &= \vec{v} \times \vec{w} \\ &= \langle -12, 18, -3\rangle \end{align}
$II$. Let $\vec{r_0} = \langle 2, 3, -5\rangle$ be the position vector of $P_0$. Then $\vec{r_0} - \vec{w} = \langle 0, 0, -15\rangle$ is the vector describing the line $\overline{P_0P_1}$, with $P_1 = (2, 3, 10) \in S$.
$III$. A vector equation for $S$ is then given by
$$\vec{n}\vec{r_0} = \vec{n}\vec{w}$$
Besides from wanting to know if my solution is correct at all, I wonder why I was given two vectors and a point to find a vector equation for $S$. It seems to me such equation could be found with the two vectors alone, by $i$. finding the normal vector of $S$ and $ii$. using the vector $\vec{v} - \vec{w}$ to describe a vector on the plane (instead of using $\vec{r_0} - \vec{w}$ as I did). Indeed, at least in my mental representation, if two vectors span a plane, a third vector defined as their difference is on the plane. Therefore I see no need for the presence of $P_0$ in this problem and I presume an equivalent solution is given by
$$ \vec{n}(\vec{v} - \vec{w}) = 0 $$ or equivalently $$ \vec{n}\vec{v} = \vec{n}\vec{w} $$
Am I confusing something here? Thanks in advance.