A method of finding the eigenvector that I don't fully understand 
Let $$A=\begin{pmatrix} a & b & c  \\ d & e & f  \\ g & h & t \\ \end{pmatrix}$$
  Which has a known eigenvalue : $\lambda$

Find the corresponding eigenvector

Over the last few months, I have come across  several ways to do this:
(1)
$$Ae=\lambda e$$
$$\begin{pmatrix} a & b & c  \\ d & e & f  \\ g & h & t \\ \end{pmatrix} \begin{pmatrix} x_1  \\ x_2  \\ x_3 \\ \end{pmatrix}= \lambda \begin{pmatrix} x_1  \\ x_2  \\ x_3 \\ \end{pmatrix}$$
Which is  the longer (more time consuming method.

(2)
Some  do it in this way:
$$(A-\lambda I)e=0$$
$$\begin{pmatrix} a-\lambda & b & c  \\ d & e-\lambda & f  \\ g & h &  t-\lambda \\ \end{pmatrix} \begin{pmatrix} x_1  \\ x_2  \\ x_3 \\ \end{pmatrix}=0$$
And mostly  they  perform reduce row echleon form to speed the process up.

(3)
Most recently I came across a method which is the fastest so far, but I don't understand how it works. And sometimes it doesn't work for me, I get correct R answer and sometimes wrong. Here is how it is, (similar to taking cross product of two vectors):
It is EVALUATING
$$\begin{bmatrix} i & j & k  \\ d & e-\lambda & f  \\ g & h & t-\lambda \\ \end{bmatrix}$$
Or
$$\begin{bmatrix} i & j & k  \\ a-\lambda & b & c  \\ d & e-\lambda & f \\ \end{bmatrix}$$
Or
$$\begin{bmatrix} i & j & k  \\ a-\lambda & b & c  \\ g & h & t-\lambda \\ \end{bmatrix}$$
Etc any two rows
How does this work? And why doesn't it work sometime? Why  doesn't it work with some matrices ? (Sometimes)
Although this is much time saving, it is unreliable to me which I believe is because I don't understand well in order to apply it correctly

Example :
Let 
$$A=\begin{pmatrix} 4 & 1 & -1  \\ -4 & -1 & 4  \\ 0 & -1 & 5 \\ \end{pmatrix}$$

$\lambda=3$

$$\begin{bmatrix} i & j & k  \\ 4-3 & 1 & -1  \\ 0 & -1 & 5-3 \\ \end{bmatrix}$$
Eigenvector:
$$\begin{pmatrix} 1   \\  -2  \\  -1 \\ \end{pmatrix}$$
But yes the evaluation goes like this.

It looks like you are computing the cross product of two rows of the matrix from the second method. This will give a vector that is orthogonal to those two rows. An eigenvector will be orthogonal to all rows, since it is in the null space of the matrix. 
So, if the two chosen vectors span the row-space, the third method gives an eigenvector.
