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Find third vector to build a base of $\mathbb{R}^3$
For the following vectors $v_1 = \left(\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}\right)$ and $v_2 = \left(\begin{matrix} 2 \\ 2 \\ 5 \end{matrix}\right)$, find a third vector $v_3 = \left(\begin{matrix} x \\ y \\ z \end{matrix}\right)$ which together build a base for $\mathbb{R}^3$.*
My solution
We know that the vectors must be linearly independent, so
$\left(\begin{matrix} 1 & 2 & x \\ 1 & 2 & y \\ 2 & 5 & z \end{matrix}\right) \left(\begin{matrix} {\lambda}_1 \\ {\lambda}_2 \\ {\lambda}_3 \end{matrix}\right) = \left(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}\right) $
for ${\lambda}_1 = {\lambda}_2 = {\lambda}_3 = 0$
The gauss reduction gives
$ \left(\begin{matrix} 1 & 2 & x \\ 0 & 1 & z-2x \\ 0 & 0 & y-x \end{matrix}\right) $
So the solution would be any vector $v_3$ for which the following hold:
$ y \neq x \\ z \neq 2x \\ x \neq 0 $
for example $ \left(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\right) $
Is this a valid solution?