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 $a$.
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
Let's define a map $\varphi : \mathbb{B} \rightarrow \mathbb{B}$ such that for every $A \in \mathbb{B}$, you obtain $\varphi(A)$ by swapping the two first rows of $A$. It is obvious that $\varphi$ is a bijection (it is an involution), so $$ \sum_{A \in \mathbb{B}} \det(A) = \sum_{A \in \mathbb{B}} \det(\varphi(A))= - \sum_{A \in \mathbb{B}} \det(A)$$
So $$\boxed{ \sum_{A \in \mathbb{B}} \det(A)=0}$$