Prove set is basis for $R^3$ Hints only

Let $P = \text{Span } \{v_1, v_2\}$ be a plane in $\mathbb{R}^3$ with normal vector $n$. Show that $\{ v_1, v_2, n\}$ is a basis for $\mathbb{R}^3$. Hints only

Equation for $P$: $P = c_1v_1 + c_2v_2$. For real $c_1, c_2$. We have by definition, $n = v_1 \times v_2$.
To make sure $\{ v_1, v_2, n\}$ is a basis for $\mathbb{R}^3$. we must have


*

*Span $\{ v_1, v_2, n\}$  =  $\mathbb{R}^3$

*$\{ v_1, v_2, n\}$ Linearly independent.


I am having major trouble showing (1).
Can I get a SMALL hint?
I am not allowed to use dimension.
Showing set is LI
By definition of a plane, $\{v_1, v_2 \}$ is linearly independent thus $\overrightarrow{0} \not \in \{v_1, v_2 \}$
So $c_1v_1 + c_2v_2 + 0 \overrightarrow{n} = 0$ Thus $\{v_1, v_2, n \}$ is LI. 
The span part is confusing
 A: You want to show that $\{ v_1, v_2, n\}$ is a basis, meaning it is a linearly-independent set generating all of $\mathbb{R}^3$.
Linear independency means that you need to show that the only way to get the zero vector is by the null linear combination. 
Now what is left is to show the generating part meaning: span$\{ v_1, v_2, n\}=\mathbb{R}^3$.
Obviously, span$\{ v_1, v_2, n\}\subseteq \mathbb{R}^3$ by definition of span and definition of a vector space.
All you need to show is that $\mathbb{R}^3\subseteq$ span$\{ v_1, v_2, n\}$. That would mean that an arbitrary vector in $\mathbb{R}^3$ can be written as a linear combination of $v_1$, $v_2$ and $n$.
This is the long way to prove it (recommended to understand). Once you are familiar with this, the fast way is to use the fact that a span is a vector subspace and a lemma that states if a vector subspace $W\subseteq V$ has the same dimension as $V$ then $W=V$.
A: Hints
$1$.If the set {$v_1,v_2,...v_r$} spans a vector space of dimension $r$ then the set is linearly independent. 
$2$.Orthogonal pair of vectors are linearly independent.
$3$. For a subspace $W$ of vector space $V$, $dim(W)=dim(V)\implies W=V$.
A: Have a look at the Gram-determinant of the three vectors, see here: https://en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant. It's clearly not zero.
Edit: You may alternatively show for any vector $w$ that
$$w-\frac{\langle n,w\rangle}{\|n\|^2}n\in\mathrm{Span}\{v_1,v_2\}.$$
