How was row reduction used to obtain the result in this example? In the example below, I obtained a different result for $(2)$, and don't know how to reproduce the given one. Here is my result:
$$ \begin{bmatrix}a \\ b \\ c \\d\end{bmatrix} = \begin{bmatrix} -2 + 2r \\ -2 + r \\ -1 \\ r \end{bmatrix}$$
I verified my result with an online row reduction calculator and it seems to be correct. 
How do we get from $(1)$ to $(2)$ by row reduction in the example below? 


Note that in your calculation, you use the last variable $d$ as the "free parameter" (you have the equation $d = r$ and then you express $a,b,c$ in terms of $r$). However, in the calculation attached from the book, they use the first variable $a$ as the free parameter. To convert your answer to the book's answer, starting with your representation of the solution, note that
$$ a = -2 + 2r = -2 + 2d \implies d = \frac{a + 2}{2} = \frac{a}{2} + 1,\\
b = -2 + r = -2 + d \implies b = -2 + \frac{a}{2} + 1 = \frac{a}{2} - 1.$$
Hence, if we use $a$ as a free parameter instead of $d$, (setting $a = r$), we get
$$ b = \frac{r}{2} - 1, \,\,\, d = \frac{r}{2} + 1. $$
It is worth emphasizing there is "nothing wrong" with your solution. It is perfectly correct and there is no need to convert it to the book's solution. Plugging your solution into the first equation, we get
$$ \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + (2r - 2) \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + (r - 2) \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + r \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$$
which is just a different parametric representation of the same line as the one written in the book's answer (verify this!).
