How to find the correct orthogonal eigenvectors for a basis of an eigenvalue? I have an exercise where I found the two correct eigenvalues $\lambda_{1} = 0$ and 
$\lambda_{2} = 6$.
The algebraic multiplicity of $\lambda_{2}$ is 2. Now I try to find $E_{\lambda_{2}}$ with the following matrix. I've already done the elementary operations needed to solve the system.
$(A-6I | 0) = \begin{pmatrix}
1 & 1 & -2 \\
0 & 0 & 0  \\
0 & 0 & 0  \\
 \end{pmatrix}$
Now we get the following system of linear equations.
$x+y-2z = 0$
$y = y$ (first free variable)
$z = z$ (second free variable)
This basically implies that $x = -y +2z$
So, to get my eigenvectors of my basis I do:
$\begin{pmatrix} 
 x \\y\\z
 \end{pmatrix} = \begin{pmatrix}-1\\1\\0 \end{pmatrix}y + 
\begin{pmatrix} 2\\0\\1 \end{pmatrix}z$
therefore $E_{\lambda_{2}} = \bigg\{ \begin{pmatrix}-1\\1\\0 \end{pmatrix} ,
\begin{pmatrix} 2\\0\\1 \end{pmatrix} \bigg\}$. 
However, this not what the solution in my textbook is even though we had the same exact matrix.
They have as solution that $E_{\lambda_{2}} = \bigg\{ \begin{pmatrix}1\\-1\\0 \end{pmatrix} ,
\begin{pmatrix} 1\\1\\1 \end{pmatrix} \bigg\}$. 
I know that they are different vectors which you can take by multiplying the vectors found in the linear equation $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = ()y + ()z$ by some scalar. 
For example I could do 
$-1\begin{pmatrix} -1\\1\\0 \end{pmatrix}$ = $\begin{pmatrix} 1\\-1\\0 \end{pmatrix}$ which gives me the first vector they had in their solution.
However, I don't get how they got $\begin{pmatrix} 1\\1\\1 \end{pmatrix}$ as their second vector of the basis. Where did I went wrong?
While writing the question I realized that in the end both my solution and theirs are correct however their solution has the advantage of proposing two orthogonal vectors that are orthogonal with the basis of $E_{\lambda_{1}} = \bigg\{\begin{pmatrix} 1\\1\\-2 \end{pmatrix}\bigg\} $ without resorting to doing Gram-Schmidt afterwards. 
So now, I want to know what was the method used to get those orthogonal vectors easily is there a process or was it just trial and error?

If you just want a vector that satisfies $x=-y+2z$ and is orthogonal to $\begin{pmatrix}-1\\1\\0\end{pmatrix}$, then solve $x=-y+2z$ and (from the dot product being $0$) $-x+y=0$. You'll find that $x=y=z$ and then might as well pick $\begin{pmatrix}1\\1\\1\end{pmatrix}$. As you mentioned, if you wanted an orthonormal basis, Gram-Schmidt is a process you could follow.
