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Extending a Linearly Independent Set of Vectors to a Basis for a Vector Space

Suppose we have a set of two linearly independent vectors  
\[\left\{ \mathbf{v_1} , \mathbf{v_2} \right\}\]
  in  
\[\mathbb{R}^3\]
, so that the only solution to  
\[\alpha \mathbf{v_1} + \beta \mathbf{v_2} =0\]
  is  
\[\alpha = \beta =0\]
.
Two vectors cannot be a basis for a three dimensional space, We can add more vectors which are not combinations of these vectors to get a spanning space, but if this expanded set is to be a basis, it must also be linearly independent.
Examnple:  
\[\left\{ \begin{pmatrix}1\\1\\2\end{pmatrix} , \begin{pmatrix}-1\\1\\1\end{pmatrix} , \mathbf{v} \right\} \]
  is to be a basis for  
\[\mathbb{R}^3\]
.
The first two vectors are linearly independent so thatbthe only solution to
\[\alpha \begin{pmatrix}1\\1\\2\end{pmatrix} + \beta \begin{pmatrix}-1\\1\\1\end{pmatrix} =0\]
  is  
\[\alpha =\beta =0 \]
.
We have to solve  
\[\alpha \begin{pmatrix}1\\1\\2\end{pmatrix} + \beta \begin{pmatrix}-1\\1\\1\end{pmatrix} + \gamma \begin{pmatrix}a\\b\\c\end{pmatrix} =0\]
  for  
\[a,b,c\]
.such that the only solution is  
\[\alpha = \beta = \gamma =0\]
.
We have
\[\begin{equation} \begin{aligned} \alpha - \beta + \gamma a &= 0 \\ \alpha + \beta + \gamma b &= 0 \\ 2 \alpha + \beta + \gamma c &=0 \end{aligned} \end{equation}\]
.
We can put  
\[a=b=0\]
  in the first two equations then  
\[\alpha =\beta =0 \]
as before.
In the thrid equation we can put  
\[\gamma =0,   c=1 \]
  to give the basis  
\[\left\{ \begin{pmatrix}1\\1\\2\end{pmatrix} , \begin{pmatrix}-1\\1\\1\end{pmatrix} , \begin{pmatrix}0\\0\\1\end{pmatrix} \right\}\]
.