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 determinant is an alternating multilinear function on the rows (or the columns, as you wish) of the matrix. What interests us here is the multilinearity.
So here, $ \det(U) = \det(R_1, R_2, R_3)$.
The matrix $U'$ obtained as you propose has determinant $\det(\frac{R_1}a, \frac{R_2}b, \frac{R_3}c)$, ie, by multilinearity, $$\begin{align*}\det(U') &= \frac 1a \det(R_1, \frac{R_2}b, \frac{R_3}c)\\ &= \frac 1{ab} \det(R_1, R_2, \frac{R_3}c)\\ &= \frac 1{abc} \det(R_1, R_2, R_3)\\& = \frac 1{abc} \det(U)\end{align*}$$
As @quid mentions in his comment, row operations are about adding a multiple of a row to another row. This operation does not affect the determinant because of its alternating and multilinear behaviour:
$\det(R_1, R_2 + a R_1, R_3) = \det(R_1, R_2, R_3) + a\det(R_1, R_1, R_3)$ by linearity, and since it is alternating, $\det(R_1, R_1, R_3) = 0$ so $\det(R_1, R_2 + a R_1, R_3) = \det(R_1, R_2, R_3)$
Edit: Geometrically speaking, the determinant is the hypervolume of the $n^{th}$-dimensional parallelogram generated by the rows of your matrix, taken as vectors. So, if your matrix is not invertible, ie its rows are linearly dependent, then the rows form a parallelogram of volume $0$, and so on.
For instance, for a $2\times2$ matrix, the determinant is the volume (=surface) of the parallelogram generated by your two rows. If these rows are linearly dependent, you can see that your parallelogram is flat, so has a volume (=surface) of $0$.
Dividing a row by $a$ means, for the determinant, dividing one of the lengths of your parallelogram by $a$, hence dividing its volume by $a$.
Best Answer
Yes that's correct, indeed by the properties of the determinant we have that
$$\det\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=\det\begin{bmatrix}a&b&c\\d&e&f\\g-a&h-b&i-c\end{bmatrix}=\\=\frac13 \det\begin{bmatrix}3a&3b&3c\\d&e&f\\g-a&h-b&i-c\end{bmatrix}=-\frac13 \det\begin{bmatrix}3a&3b&3c\\-d&-e&-f\\g-a&h-b&i-c\end{bmatrix}$$