Sure you can get more than one answer! A right one and any number of wrong ones!
Here, you are row reducing the matrix
\(\displaystyle \begin{pmatrix}1 & 1 & -1 \\ 2 & 5 & 3 \\ 1 & 10 & -11\end{pmatrix}\)
and you start off by subtracting twice the first row from the secon row and subtracting the first row from the third:
\(\displaystyle \begin{pmatrix}1 & 1 & -1 \\ 0 & -3 & 5 \\ 0 & 6 & -10\end{pmatrix}\)
That is a good first step.
Now, your
objective is to get a triangular matrix with "0"s below the main diagonal, right? So I don't understand why you would divide the third row by 2. It doesn't hurt, but it does't help! Instead, to get a "0" in the second column of the third row, you need to add two times the second row to the third: 6+ 2(-3)= 0 and -10+ 2(5)= 0. That is,
\(\displaystyle \begin{pmatrix}1 & 4 & -1 \\ 0 & -3 & 5 \\ 0 & 0 & 0\end{pmatrix}\).
As I said, while dividing the third row by 2, to get
\(\displaystyle \begin{pmatrix}1 & 4 & -1 \\ 0 & -3 & 5 \\ 0 & 3& -5\end{pmatrix}\)
doesn't help, it doesn't hurt- you just haven't finished. Now add the second row to the third row to get, again,
\(\displaystyle \begin{pmatrix}1 & 4 & -1 \\ 0 & -3 & 5 \\ 0 & 0 & 0\end{pmatrix}\).