You're getting different answers because you're subtracting $R_3$ from $3 R_1$. Usually, you subtract $3 R_1$ from $R_3$ (you can add or subtract rows, but you're changing the sign on $R_3$ here: we often choose to do it this way because adding or subtracting a multiple of a row keeps the determinant the same).
So, to get Wikihow's answer, you should have taken $R_3 - 3R_1 \to R_3$ instead. This also explains the sign flip, since $R_3 - 3R_1 = - ( 3R_1 - R_3 )$.
"For what value of $k$ does the system not have a unique solution"
This would include both the case of infinitely many solutions and also the case of no solution.
A much easier approach than performing row operations is the following:
Let us express this as a matrix equation $Ax=b$ instead of as a system of equations. There is a theorem:
$Ax=b$ has a unique solution if and only if $A$ is invertible.
There is another theorem:
$A$ is invertible if and only if $\det(A)\neq 0$
So, the value(s) for $k$ such that the system does not have a unique solution are those such that $\det(A)=0$.
$\det(A)=1(k\cdot 3 - (-1)\cdot(-1)) - (-2)(2\cdot 3 - (-1)\cdot 1) + 2(2\cdot (-1) - k\cdot 1)$
$=(3k-1)+(14)+(-4-2k)$
$= k+9$
So, if $\det(A)=0$, you would have $k+9=0$, which implies that $k=-9$
As for your method and current work, you are nearly there. You left off at:
$=\left[\begin{array}{ccc|c}1&-2&2&0\\0&0&4k+36&-92-20k\\0&4k&4k&-20k\end{array}\right]$
Now, rowswap to get $R_2\leftrightarrow R_3$
$=\left[\begin{array}{ccc|c}1&-2&2&0\\0&4k&4k&-20k\\0&0&4k+36&-92-20k\end{array}\right]$
It will be that there are no solutions when the final row looks like $[0~0~0~|~n]$ with $n\neq 0$, or infinitely many solutions if the final row looks like $[0~0~0~|~0]$ (assuming that there are no other $[0~0~0~|~n]$ rows elsewhere after row reduction is complete)
To get the third entry of the last row equal to zero, that would happen when $4k+36=0$ which occurs when $k=-9$.
There is a slight error in your work which is difficult to notice, and that is the step where you multiplied a row by $k$. In the case that you allow $k=0$, you have in effect destroyed the information that that row could have given you. Remember that you are only allowed to multiply rows by nonzero numbers. When multiplying by unknown quantities, you should be careful after reaching a final answer that doing so didn't cause a problem (like it did this time). Your work would imply that by setting $k=0$ the second row becomes all zeroes, implying infinitely many solutions, but that is not the case (as shown by my first method).
Instead, of doing $4R_2\mapsto R_2$, followed by $kR_3\mapsto R_3$, followed by $R_2+R_3\mapsto R_2$., you would have avoided this error by simply doing $4R_2+kR_3\mapsto R_3$. Doing so wouldn't have caused any loss of information.
Best Answer
The reduced row echelon form is unique. The are many paths to this form, but the destination is unique.
Let's look at the example problem. Use augmented reduction and clear the columns sequentially.
Column 1 $$ % E \left[ \begin{array}{rcc} \frac{1}{2} & 0 & 0 \\ -\frac{1}{2} & 1 & 0 \\ -\frac{1}{2} & 0 & 1 \\ \end{array} \right] % A \left[ \begin{array}{cccrc|ccc} 2 & 1 & 4 & -4 & 11 & 1 & 0 & 0 \\ 1 & 2 & 3 & 1 & 8 & 0 & 1 & 0 \\ 1 & 1 & 2 & -1 & 6 & 0 & 0 & 1 \\ \end{array} \right] % = % \left[ \begin{array}{cccrc|rcc} \boxed{1} & \frac{1}{2} & 2 & -2 & \frac{11}{2} & \frac{1}{2} & 0 & 0 \\ 0 & \frac{3}{2} & 1 & 3 & \frac{5}{2} & -\frac{1}{2} & 1 & 0 \\ 0 & \frac{1}{2} & 0 & 1 & \frac{1}{2} & -\frac{1}{2} & 0 & 1 \\ \end{array} \right] % $$
Column 2 $$ % E \left[ \begin{array}{crc} 1 & -\frac{1}{3} & 0 \\ 0 & \frac{2}{3} & 0 \\ 0 & -\frac{1}{3} & 1 \\ \end{array} \right] % A \left[ \begin{array}{cccrc|rcc} \boxed{1} & \frac{1}{2} & 2 & -2 & \frac{11}{2} & \frac{1}{2} & 0 & 0 \\ 0 & \frac{3}{2} & 1 & 3 & \frac{5}{2} & -\frac{1}{2} & 1 & 0 \\ 0 & \frac{1}{2} & 0 & 1 & \frac{1}{2} & -\frac{1}{2} & 0 & 1 \\ \end{array} \right] % = % \left[ \begin{array}{ccrrc|rrc} \boxed{1} & 0 & \frac{5}{3} & -3 & \frac{14}{3} & \frac{2}{3} & -\frac{1}{3} & 0 \\ 0 & \boxed{1} & \frac{2}{3} & 2 & \frac{5}{3} & -\frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & -\frac{1}{3} & 0 & -\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} & 1 \end{array} \right] % $$
Column 3 $$ % E \left[ \begin{array}{ccr} 1 & 0 & 5 \\ 0 & 1 & 2 \\ 0 & 0 & -3 \\ \end{array} \right] % A \left[ \begin{array}{ccrrc|rrc} \boxed{1} & 0 & \frac{5}{3} & -3 & \frac{14}{3} & \frac{2}{3} & -\frac{1}{3} & 0 \\ 0 & \boxed{1} & \frac{2}{3} & 2 & \frac{5}{3} & -\frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & -\frac{1}{3} & 0 & -\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} & 1 \end{array} \right] % = % \left[ \begin{array}{cccrc|rrr} \boxed{1} & 0 & 0 & -3 & 3 & -1 & -2 & 5 \\ 0 & \boxed{1} & 0 & 2 & 1 & -1 & 0 & 2 \\ 0 & 0 & \boxed{1} & 0 & 1 & 1 & 1 & -3 \\ \end{array} \right] % $$