Prove that a subset $E$ of $\mathbb{R}^3$ is exactly a plane, if there exists vectors $u,v,w \in \mathbb{R}^3$, such that $v$ and $w$ are linearly independent and
$$E = u + \mathbb{R}v + \mathbb{R}w$$
Hint: Let $v,w \in \mathbb{R}^3$ be two linearly independent vectors. It can be used that a vector $x \in \mathbb{R}^3$ is perpendicular to the Cross Product
$$ v \times w := \begin{pmatrix} v_2w_3 - v_3w_2, \\ v_3w_1 - v_1w_3, \\ v_1w_2 - v_2w_1 \end{pmatrix}^T$$
exactly if $x = \lambda v + \mu w$ for some $\lambda, \mu \in \mathbb{R}$
There are several equation that can describe a plane
For example $$n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0$$ where $n = (n_1, n_2, n_3)$ is a vector normal (perpendicular) to the plane, and $(x_0, y_0, z_0)$ is a point on the plane.
Then there is also the definition above $$r = r_0 + s \cdot v + t \cdot w$$ where $r_0$ is a point on the plane, and $v$ and $w$ are two linearly independent vectors that span the plane.
In our case, the cross product in the hint obviously gives us a perpendicular vector, so we can find $n = (n_1, n_2, n_3)$. The vector $x$ in the hint is assumed to be perpendicular to the Cross Product, which means it lies in the plane. But how does that help ? What should I do next ?