I found the following in my notes:
The plane that is constructed by two non-parallel vectors $\overrightarrow{v}$ and $\overrightarrow{w}$ consists of all the points of the form $a \overrightarrow{v}+b\overrightarrow{w}$, $a, b \in \mathbb{R}$.
The plane that is defined by $\overrightarrow{v}$ and $\overrightarrow{w}$ is called the plane that is produced by $\overrightarrow{v}$ and $\overrightarrow{w}$.
If $\overrightarrow{v}$ is a multiple of $\overrightarrow{w}$ and $\overrightarrow{w} \neq 0$, then $\overrightarrow{v}$ and $\overrightarrow{w}$ are parallel.
I am asked to find the plane that is produced by the two vectors $\overrightarrow{v_1}=(3, 8, 0)$ and $\overrightarrow{v_2}=(0, 3, 8)$.
Is the plane $a \overrightarrow{v_1}+b\overrightarrow{v_2}$ ?? Or is this only the form of points of the plane??
There are several common descriptions of a 3D plane: $$ a_1 v_1 + a_2 v_2 \quad (a_1, a_2 \in \mathbb{R}) $$ The above one is the one you assumed in your question.
Another way to model all vectors $x$ of a 3D plane is by a normal vector $n$: $$ 0 = x \cdot n = a_1 (v_1 \cdot n) + a_2 (v_2 \cdot n) $$ so $n$ should fulfill these two conditions: \begin{align} 0 &= (3,8,0) \cdot (n_1, n_2, n_3) = 3 n_1 + 8 n_2 \wedge \\ 0 &= (0,3,8) \cdot (n_1, n_2, n_3) = 3 n_2 + 8 n_3 \end{align}
You might guess the normal vector or use the vector product to derive one: $$ n = v_1 \times v_2 = (8\cdot 8 - 0\cdot 3, 0\cdot 0 - 3\cdot 8, 3 \cdot 3 - 8 \cdot 0) = (64, -24, 9) $$