What am I doing wrong solving this system of equations?

$$\begin{cases} 2x_1+5x_2-8x_3=8\\ 4x_1+3x_2-9x_3=9\\ 2x_1+3x_2-5x_3=7\\ x_1+8x_2-7x_3=12 \end{cases}$$

From my elementary row operations, I get that it has no solution. (Row operations are to be read from top to bottom.)

$$\left[\begin{array}{ccc|c} 2 & 5 & -8 & 8 \\ 4 & 3 & -9 & 9 \\ 2 & 3 & -5 & 7 \\ 1 & 8 & -7 & 12 \end{array}\right] \overset{\overset{\large{R_1\to R_1-R_3}}{{R_2\to R_2-2R_3}}}{\overset{R_3\to R_3-2R_4}{\large\longrightarrow}} \left[\begin{array}{ccc|c} 0 & 2 & -3 & 1 \\ 0 & -3 & 1 & -5 \\ 0 & -13 & 9 & -17 \\ 1 & 8 & -7 & 12 \end{array}\right] \overset{\overset{\large{R_3\,\leftrightarrow\, R_4}}{R_2\,\leftrightarrow\, R_3}}{\overset{R_1\,\leftrightarrow\,R_2}{\large\longrightarrow}} \left[\begin{array}{ccc|c} 1 & 8 & -7 & 12 \\ 0 & 2 & -3 & 1 \\ 0 & -3 & 1 & -5 \\ 0 & -13 & 9 & -17 \end{array}\right]$$

$$\overset{R_4\to R_4-R_3}{\large\longrightarrow} \left[\begin{array}{ccc|c} 1 & 8 & -7 & 12 \\ 0 & 2 & -3 & 1 \\ 0 & -3 & 1 & -5 \\ 0 & 10 & 8 & -12 \end{array}\right] \overset{\overset{\large{R_3\to R_3+R_2}}{R_4\to R_4-5R_2}}{\large\longrightarrow} \left[\begin{array}{ccc|c} 1 & 8 & -7 & 12 \\ 0 & 2 & -3 & 1 \\ 0 & -1 & -2 & -4 \\ 0 & 0 & 23 & -17 \end{array}\right] \overset{\overset{\large{R_2\to R_2+2R_3}}{R_3\to-R_3}}{\large\longrightarrow}$$

$$\left[\begin{array}{ccc|c} 1 & 8 & -7 & 12 \\ 0 & 0 & -7 & -7 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & 23 & -17 \\ \end{array}\right] \overset{R_2\,\leftrightarrow\,R_3}{\large\longrightarrow} \left[\begin{array}{ccc|c} 1 & 8 & -7 & 12 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & -7 & -7 \\ 0 & 0 & 23 & -17 \\ \end{array}\right]$$

However, the answer in the book $(3, 2, 1)$ fits the system.

Was there an arithmetical mistake, or do I misunderstand something fundamentally?

Hint: Try inputing the solution $(3,2,1)$ into every step. That will allow you to identify the step where you went wrong.

You do (in the third matrix): $$L3-L4=(0, -3, 1 \mid -5)-(0, -13, 9 \mid -19)=(0, 10, -8 \mid 12)$$ but you have $(0, 10, 8 \mid -12)$ instead.