The final matrix should be
$$
\begin{bmatrix}
3 & 3 & -3 \\
0 & -1 & 1 \\
0 & 0 & 24
\end{bmatrix}
$$
However, you have multiplied the determinant by $-1$ with the first operation and by $-3$ with the second one, so you get
$$
\frac{3\cdot(-1)\cdot24}{(-1)\cdot(-3)}=-24
$$
I use a different method, reducing the pivots to $1$:
\begin{align}
\begin{bmatrix}
3 & 3 & -3 \\
3 & 4 & -4 \\
2 & -3 & -5
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & -1 \\
3 & 4 & -4 \\
2 & -3 & -5
\end{bmatrix} && R_1\gets\color{red}{\frac{1}{3}}R_1
\\[6px]
&\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
2 & -3 & -5
\end{bmatrix} && R_2\gets R_2-R_1
\\[6px]
&\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & -5 & -3
\end{bmatrix} && R_3\gets R_3-2R_1
\\[6px]
&\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & 0 & -8
\end{bmatrix} && R_3\gets R_3+5R_2
\\[6px]
&\to
\begin{bmatrix}
1 & 1 & -1 \\
0 & 1 & -1 \\
0 & 0 & 1
\end{bmatrix} && R_3\gets \color{red}{-\frac{1}{8}}R_3
\end{align}
The red numbers tell that the determinant has been multiplied by
$$
-\frac{1}{24}
$$
so it is $-24$.
The key idea in using row operations to evaluate the determinant of a matrix is the fact that a triangular matrix (one with all zeros below the main diagonal) has a determinant equal to the product of the numbers on the main diagonal. Therefore one would like to use row operations to 'reduce' the matrix to triangular form.
However, the effect of using the three row operations on a determinant are a bit different than when they are used to reduce a system of linear equations.
(1) Swapping two rows changes the sign of the determinant
(2) When dividing a row by a constant, the constant becomes a factor written in front of the determinant.
(3) Adding a multiple of one row to another does not change the value of the determinant.
Let's apply these operations to your matrix to find its determinant.
First we want to produce two zeros in rows $2$ and $3$ of column $1$. (Remember our goal is to produce all zeros below the main diagonal, and we do this one column at a time beginning with column $1$.)
The two row operations $-R_1+R_2\to R_2$ and $-4R_1+R_3\to R_3$ will accomplish this goal, and will not change the value of the determinant.
\begin{eqnarray}
\begin{vmatrix}1 & 7 & -3\\ 1 & 3 & 1 &\\ 4 & 8 & 1 \end{vmatrix} &=&
\begin{vmatrix} 1 & 7 & -3\\0 & -4 & 4\\0 & -20 & 13\end{vmatrix}
\end{eqnarray}
Now all that remains is to obtain a $0$ in row $3$ column $2$. We see that adding $-5$ times row $2$ to row $3$ will accomplish this. That is, $-5R_2+R_3\to R_3$.
\begin{eqnarray}
\begin{vmatrix}1 & 7 & -3\\ 1 & 3 & 1 &\\ 4 & 8 & 1 \end{vmatrix} &=&
\begin{vmatrix} 1 & 7 & -3\\0 & -4 & 4\\0 & -20 & 13\end{vmatrix}\\
&=& \begin{vmatrix}
1 & 7 & -3\\0 & -4 & 4\\0 & 0 & -7
\end{vmatrix}
\end{eqnarray}
Since we only had to use the third row operation, the one which does not change the value of the determinant and since we now have a triangular matrix, we find the determinant by multiplying the numbers on the main diagonal.
\begin{eqnarray}
\begin{vmatrix}1 & 7 & -3\\ 1 & 3 & 1 &\\ 4 & 8 & 1 \end{vmatrix} &=&
\begin{vmatrix} 1 & 7 & -3\\0 & -4 & 4\\0 & -20 & 13\end{vmatrix}\\
&=& \begin{vmatrix}
1 & 7 & -3\\0 & -4 & 4\\0 & 0 & -7
\end{vmatrix}\\
&=&28
\end{eqnarray}
Best Answer
Recall the effect of elementary column (or row) operations on the determinant:
Swapping two columns changes the sign of the determinant.
Multiplying a column by a nonzero constant results in the determinant being multiplied by that constant.
Adding a multiple of a column to another column does not change the determinant.
You described operations 1 and 2, but you are missing the third which in this case would be adding $6$ times the first column to the last column. This will get us to arrive at the matrix $A$. In each of these elementary column operations, do not forget the effect on the value of the determinant as stated above.
Therefore, the operations are $-\dfrac{1}{9}C_1$, $\dfrac{1}{8}C_2$, $R_1\leftrightarrow R_2$, and $6C_1+C_3\rightarrow C_3$, where $R_i$ and $C_i$ indicate the rows and columns, respectively.
In the end, you should get $\det B=9(8)(-4)=-288$.