$$A=\begin{bmatrix}1 & 2 & 1\\-2 & a & 0 \\ 2 & 10 & a+2\end{bmatrix},a\in\Bbb{R}$$
it's not allowed to use determinants.
In the next step I get that:
$$A=\begin{bmatrix}1 & 2 & 1\\0 & a+2 & 2 \\ 0 & 6 & a\end{bmatrix}$$
and in the next step:
if $a=0$ then
$$A=\begin{bmatrix}1 & 2 & 1\\0 & 2 & 2 \\ 0 & 6 & 0\end{bmatrix}$$ so the rank = $2$
if $ a = -2$
$$A=\begin{bmatrix}1 & 2 & 1\\0 & 6 & -2 \\ 0 & 0 & -2\end{bmatrix}$$ which is equal to
$$A=\begin{bmatrix}1 & 2 & 1\\0 & 6 & -2 \\ 0 & 0 & 2\end{bmatrix}$$
so the rank is = $3$
if $a$= something else:
$$A=\begin{bmatrix}1 & 2 & 1\\0 & a+2 & 2 \\ 0 & 6 & a\end{bmatrix}$$
so the rank is =$3$
Is my solution in this form correct? I appreciate any kind of help a lot.
You made a mistake in the first step, it should be
$$A=\begin{bmatrix}1 & 2 & 1\\0 & a+\color{red}4 & 2 \\ 0 & 6 & a\end{bmatrix}$$
To proceed, you might like to interchange the second row and the third row and then use the second row as the pivot for the next operation.
You might then like to consider cases for $a=2$ and $a=-6$.