Beginner troubleshooting an eigenvector calculation I am having some difficulty identifying the error in my eigenvector calculation. I am trying to calculate the final eigenvector for $\lambda_3 = 1$ and am expecting the result $ X_3 = \left(\begin{smallmatrix}-2\\17\\7\end{smallmatrix}\right)$
To begin with, I set up the following equation (for the purpose of this question I will refer to the leftmost matrix here as A). 
$$
\begin{bmatrix}
   1 - \lambda & 0 & 0 \\ 
   3 & 3 - \lambda  & -4\\ 
   -2 & 1 & -\lambda -2 \\
\end{bmatrix}
\begin{bmatrix}
  x_1 \\ 
   x_2\\ 
   x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
0\\ 
0\\ 
0\\
\end{bmatrix}
$$
I) Substitute $\lambda_3 = 1$
$$
\begin{bmatrix}
   0 & 0 & 0 \\ 
   3 & 2  & -4\\ 
   -2 & 1 & -3 \\
\end{bmatrix}
\begin{bmatrix}
  x_1 \\ 
   x_2\\ 
   x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
0\\ 
0\\ 
0\\
\end{bmatrix}
$$
II) Reduce the matrix with elementary row operations.
$R_2 \leftarrow R_2 - 2R_3$
$$
A = 
\begin{bmatrix}
   0 & 0 & 0 \\ 
   7 & 0  & 2\\ 
   -2 & 1 & -3 \\
\end{bmatrix}
$$
$R_3 \leftarrow  3R_2 +  2R_3$
$$
A = 
\begin{bmatrix}
   0 & 0 & 0 \\ 
   7 & 0  & 2\\ 
   17 & 2 & 0 \\
\end{bmatrix}
$$
$R_2 \leftarrow  \frac{1}{7} R_2$
$R_3 \leftarrow \frac{1}{17} R_3$$
$$
A = 
\begin{bmatrix}
   0 & 0 & 0 \\ 
   1 & 0  & 2/7\\ 
   1 & 2/17 & 0 \\
\end{bmatrix}
$$
III) multiply matrices to get a series of equations equal to 0 and rearrange them in terms of a common element.
$x_1 + \frac{2}{7}x_3 = 0 \rightarrow x_1 = -\frac{2}{7}x_3$
$x_1 + \frac{2}{17}x_2 = 0 \rightarrow x_1 = -\frac{2}{17}x_2$
IV) Substitute a value into the vector to get an eigenvector.
Let $\ x_1 = 1 \rightarrow X_3 =  \left(\begin{smallmatrix}1\\-2/17\\-2/7\end{smallmatrix}\right)
$
Which at this point we can see is not a multiple of the expected $X_3$. Can anyone highlight my error for me?
Many thanks in advance.

You have solved the problem correctly up to $$x_1 + \frac{2}{7}x_3 = 0 \rightarrow x_1 = -\frac{2}{7}x_3$$ and $$x_1 + \frac{2}{17}x_2 = 0 \rightarrow x_1 = -\frac{2}{17}x_2$$ Now when you let $x_1=1$ you get $x_3=-\frac {7}{2}$ and $x_2=-\frac {17}{2}$ which give you the correct eigenvector.  
