# Finding a vector equation of the plane determined by $3x+3y+z=1$

I'm preparing for an upcoming test and wanted to validate my solution to the following problem:

Find a vector equation of the plane determined by $$3x+3y+z=1$$

Here's what I did

$$I$$. Let $$x=y=1$$. Then from $$3x+3y+z=1$$ it follows that $$\vec{r_0}=[1, 1, -5]$$ is the position vector of a point in the plane.

$$II$$. From the cartesian equation it follows that $$\vec{n}=[3, 3, 1]$$ is a vector normal to the plane. Then for any vectors $$\vec{v}, \vec{w}$$ such that $$\vec{v} \neq t \vec{w}$$ (linearly independent), if $$\vec{v}$$ and $$\vec{w}$$ are normal to $$\vec{n}$$ they are direction vectors of two lines (or vectors) in the plane. Notice that

\begin{align} \vec{n} \cdot \vec{a}=0 &\iff 3a_1+3a_2+a_3=0\end{align}

If $$a_1=a_2$$, then $$-6a_1=a_3$$ is a solution. If $$a_1=a_3$$ then $$-\frac{4}{3}a_1=a_2$$ is a solution. We can then define

$$\vec{v}:=[1, 1, -6], \vec{w}:=[1, -\frac{4}{3}, 1]$$

knowin both $$\vec{v}$$ and $$\vec{w}$$ are normal to $$\vec{n}$$ (this is easy to verify algebraically) and therefore direction vectors for two lines (or vectors) in the plane.

$$III$$. This is enough to give

\begin{align} \vec{g}(t, s)&=\vec{r}_0 + t\vec{v}+s\vec{w} \\\vec{g}(t, s)&=[1, 1, -5] + t[1, 1, -6]+s[1, -\frac{4}{3}, 1]\end{align}

as a vector equation of the plane.

Is this solution correct? If not, why? If so, were there simpler approaches to the problem? Thanks in advance.

An alternative way is to find three non-collinear points in the plane, $$A, B$$ and $$C$$, and then you can construct the vector equation of the plane as $$\underline{r}=\overrightarrow{OA}+\lambda\overrightarrow{AB}+\mu\overrightarrow{AC}$$