Find an equation of the plane spanned by $v_1$ and $v_2$, find a vector $v_3$ that can be added to produce a basis for $\mathbb{R}^3$ Let $v_1=(-1,2,3)$ and $v_2=(5,3,-1)$. Find the equation of the plane spanned by $v_1$ and $v_2$. Also find a vector $v_3$ that can be added to the set $\{v_1,v_2\}$ to produce a basis for $\mathbb{R}^3$.
I'm stuck on both parts because everywhere I read it says I need a point in addition to these two vectors to find the equation. 
For the second part I'm not sure of the method to find a vector to produce a basis, we haven't been told how to do that... I know a basis for $\mathbb{R}^3$ would have to have $3$ vectors that span $\mathbb{R}^3$ and must be L.I., so I know that $kv_1 + kv_2 + kv_3 = 0$ must be $k_1=k_2=k_3 = 0$ ...but what is a concrete method to find a third vector $v_3$? 
Forgive me if I missed somewhere I could have found this out, I couldn't seem to find the method.
 A: To find a third vector to produce a basis for $\mathbb{R^3}$, take your vectors v1 and v2 and row reduce to find out where your pivots are. 
$$
\begin{pmatrix}
-1&5\\
2&3\\
3&-1\\
\end{pmatrix}
$$ is row equivalent to:
$$
\begin{pmatrix}
1&0\\
0&1\\
0&0\\
\end{pmatrix}
$$
Because we need a pivot in the third row to produce a basis for $\mathbb{R^3}$, we can add the vector 
$$
\begin{pmatrix}
0\\
0\\
1\\
\end{pmatrix}
$$
to produce a basis set.
A: Let $\begin{pmatrix}x\\y\\z\end{pmatrix}$ be a point in Span$\{v_1,v_2\}$, so there are $t,r\in\mathbb{R}$ such that
\begin{align*}
\begin{pmatrix}x\\y\\z\end{pmatrix}&=tv_1+rv_2\\
&=t\begin{pmatrix}-1\\2\\3\end{pmatrix}+r\begin{pmatrix}5\\3\\-1\end{pmatrix}\\
&=\begin{pmatrix}-1&5\\2&3\\3&-1\end{pmatrix}\begin{pmatrix} t \\ r\end{pmatrix}...(1)
\end{align*}
This system can be written as $$\begin{pmatrix}-1&5&|&x\\2&3&|&y\\3&-1&|&z\end{pmatrix}$$
By mean of elementary operations by rows we get $$\begin{pmatrix}1&0&|&4x-5y+5z\\0&1&|&x-y+z\\0&0&|&11x-14y+13z\end{pmatrix}$$
Then, (1) has solutions iff $11x-14y+13z=0$, this is the equation of the plane.
A: Consider general equation of the plane $ax+by+cz=d$.
The plane that passes through origin can be expressed as:
$ax+by+cz=0$ and even further, $Ax+By+z=0$.
Then first vector gives $A(-1)+B(2) + 3=0$; and the second $A(5)+B(3)-1=0$.
Solving the above system for A and B gives $A=11/13; B=-14/13$.
Hence the plane is $11x-14y+13z=0\cdot 13=0$.
Note that vector $(11;-14;13)$ is perpendicular to the plane, meaning that we can take it as $v_3$ to span $\mathbb R^3$.
