I start with $\begin{bmatrix} 1 & 2k & 1 & 0 \\ 0 & 1-6k & k-3 & 2 \\ 0 & 7k & 4 & 2 \end{bmatrix}$
I want to get it in row echelon form, so I'm looking for a zero in row 3, column 2.
I thought of multiplying row 3 by whatever I'd need to make that zero, and came up with $R_3 \implies R_3 - \frac{7}{2} R_1$
This gives $\begin{bmatrix} 1 & 2k & 1 & 0 \\ 0 & 1-6k & k-3 & 2 \\ -\frac{7}{2} & 0 & \frac{1}{2} & 2 \end{bmatrix}$
However this leaves a contradiction in the third row, implying no solution, and I know from the full worked example that there are solutions for some values of k.
Instead of my step of $R_3 \implies R_3 - \frac{7}{2} R_1$, the worked example does a fairly hairy calculation of $R_3 \implies 6R_3 + 7R_2$.
So whilst I'm prepared to believe that the worked example step is the "right" one, why was my step wrong? Somewhat confused.
There are a lot of ways to get a matrix into r.r.e.f. or r.e.f.. If the series of algebraic steps you preformed worked out, then you most likely did it correct and your step is 'valid.' But, if you find yourself stuck after preforming an algebraic process, you have either taken the hardest possible route to get the matrix to r.e.f.. or have made an error.
Your initial step could work, but by doing so, you introduced another term into the first column of your matrix. The goal is to have only one term per column. By doing $R_3-(7/2)*R_1$, you have taken a step backwards. So, you will have to do something more complicated in the future to correct that.