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
It turns out that$$\begin{pmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{pmatrix}=\begin{pmatrix}1&a&bc\\1&b&ac\\1&c&ab\end{pmatrix}.\begin{pmatrix}1&0&-ab-ac-bc\\0&1&a+b+c\\0&0&1\end{pmatrix}.$$Since, obviously,$$\begin{vmatrix}1&0&-ab-ac-bc\\0&1&a+b+c\\0&0&1\end{vmatrix}=1,$$the two given matrices have the same determinant.