If I have to find the Eigenvalue of a 3x3 matrix of the type
$$A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & 0 \\ 0 & 0 & 7 \end{bmatrix}$$
The three eigenvalues are $\lambda_{1} = 3$, $\lambda_{2} = 1$, $\lambda_{3} = 7$
for $\lambda_{3}$ we can plug $\lambda$ into $A-\lambda I$ which gives:
$$\begin{bmatrix} -5 & -1 & 0 \\ -1 & -5 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$$
Personally I spontaneously went into the direction of getting row echelon form as follow: After having swapped the first and second row, I obtained the following matrix:
$$\begin{bmatrix} -1 & -5 & 0 \\ -5 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$
and then tried to obtain the row echelon form by substracting diving five times the first row from the second row, I obtained
$$\begin{bmatrix} -1 & -5 & 0 \\ 0 & -24 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$
However how can we conclude the following statement
Since $\det\begin{bmatrix} -5 & -1\\ -1 & -5 \end{bmatrix} = 24 \neq 0$ we know that $x = y = 0$ (and the third equations reads as $0x + 0y + 0z = 0$ i.e. we are free to choose the value of $z$.
You have made a bit of a mess of things.
If $v$ is an eigenvector then.
$Av =\lambda v\\ (A-\lambda I)v = 0$
So, you want to find the kernel of your matrix above.
But then you make an arithmetic mistake.
if $\lambda = 7$
$\begin{bmatrix} 2-\lambda& -1&0\\-1&3-\lambda &0\\0&0&7-\lambda \end{bmatrix} =\begin{bmatrix} -5& -1&0\\-1&-4 &0\\0&0&0 \end{bmatrix}$
(actually I think you mis-typed your initial matrix. As, 1,3 are not eigenvectors of the matrix posted)
Assuming you meant for you initial matrix to be
$\begin{bmatrix} 2& -1&0\\-1&2 &0\\0&0&7\end{bmatrix}$
To find the kernel of the matrix
$\begin{bmatrix} -5& -1&0\\-1&-5 &0\\0&0&0 \end{bmatrix}$
No row operations are necessary. The zeros in the right-hand column guarantee
$\begin{bmatrix} -5& -1&0\\-1&-5 &0\\0&0&0 \end{bmatrix} \begin{bmatrix} 0\\0\\1\end{bmatrix}=0$
This is the easiest of the 3 eigenvectors to find.
Since we have 3 independent eigenvalues of a $3\times 3$ matrix, the eigenspace for this eigenvalue is degree 1. Once we have found one vector we are done.
$\lambda = 1$
$\begin{bmatrix} 1& -1&0\\-1&1 &0\\0&0&6 \end{bmatrix} \begin{bmatrix} 1\\1\\0\end{bmatrix}=0$
$\lambda = 3$
$\begin{bmatrix} -1& -1&0\\-1&-1 &0\\0&0&4 \end{bmatrix} \begin{bmatrix} 1\\-1\\0\end{bmatrix}=0$