Equation of plane containing two vectors

I am struggling with the interpretation of this question:

Vectors: $u = \left(1,\ 0,\ \sqrt3 \right)$ and $v = (1,\ \sqrt3,\ 0)$ in standard position.

Find an equation of the plane containing $u$ and $v$.

Am I correct in interpreting this question that the plane is parallel to both vectors? Does this mean that I can form the cross product of $u$ and $v$ to find the normal and use one of the vectors as a point on the plane?

Thank you in advance

If you're using the cross product, this gives $$ \begin{align} \begin{vmatrix} e_{\vec{x}} & e_{\vec{y}} & e_{\vec{z}} \\ 1 & 0 & \sqrt{3} \\ 1 & \sqrt{3} & 0 \end{vmatrix} = \left(\begin{vmatrix} 0 & \sqrt{3} \\ \sqrt{3} & 0\end{vmatrix}, -\begin{vmatrix} 1 & \sqrt{3} \\ 1 & 0\end{vmatrix}, \begin{vmatrix} 1 & 0 \\ 1 & \sqrt{3}\end{vmatrix}\right) = \left(-3, \sqrt{3}, \sqrt{3} \right) \end{align} $$

Thus, $\vec{p} = t(-3,\sqrt{3},\sqrt{3}), \enspace t\neq 0$ is a orthogonal vector to vector $\vec{u}$ and $\vec{v}$ and the plane they make. If $(x,y,z) and (x_0,y_0,z_0)$ are points on the plane, then $\vec{w} = (x-x_0, y-y_0,z-z_0)$ is a vector on the plane, and $\vec{w}\cdot\vec{p} = 0$:

$$ \begin{align} t(-3,\sqrt{3},\sqrt{3})\cdot(x-x_0, y-y_0,z-z_0) &= 0 \\ t(-3(x-x_0)+\sqrt{3}(y-y_0)+\sqrt{3}(z-z_0) &= 0 \end{align} $$

We can choose $t=1$ (because a vector is orthogonal regardless of its length) and $(x_0,y_0,z_0) = (0,0,0)$ (because the plane passes through the origin, because both vectors go through the origin). This yields:

$$-3x+\sqrt{3}y+\sqrt{3}z = 0$$

Well, since $u$ and $v$ are not parallel, the plane $P$ is

$$ P = \{a \cdot u + b \cdot v \} $$

for $a, b \in \mathbb{R}$. So

$$ P = \{(a, 0, \sqrt{3}a) ~ + ~ (b, \sqrt{3}b, 0) \} = \{(a + b, \sqrt{3}b, \sqrt{3}a)\} $$

To obtain the equation of the plane in Euclidean / Cartesian / xyz-form let

$$ (a + b, \sqrt{3}b, \sqrt{3}a) = (x, y, z)$$

and observe that $z - \sqrt{3}x + y = 0$.