Solving the system produces
$$X= 1,Y= 3,Z= 5,W= 7$$
Yes, swapping rows is a legal operation and moreover, it is required for this method to converge, see: Switching Rows in the Gauss Seidel method.
We cannot put the system into diagonally dominant form (due to the second row), but we at least have the rows' dominant term in the diagonal
$$ A = \left(
\begin{array}{cccc}
6 & 1 & 1 & 1 \\
-2 & 5 & 3 & 2 \\
1 & -1 & 7 & 4 \\
-1 & 2 & -2 & 7 \\
\end{array}
\right), ~~~ b = \left(
\begin{array}{c}
21 \\
42 \\
61 \\
44 \\
\end{array}
\right)$$
Selecting a starting point of
$$x^{(1)} = \left(
\begin{array}{c}
1 \\
1 \\
1 \\
1 \\
\end{array}
\right)$$
This results of the iteration are
$$\left(
\begin{array}{cccc}
1 & 1 & 1 & 1. \\
3. & 8.6 & 8.94286 & 6.81224 \\
-0.559184 & 0.0857143 & 4.9137 & 7.58526 \\
1.40255 & 2.9787 & 4.60502 & 6.95074 \\
1.07759 & 3.28773 & 5.05817 & 6.9455 \\
0.951435 & 2.96748 & 5.03344 & 7.01191 \\
0.997863 & 2.97432 & 4.98983 & 7.00413 \\
1.00529 & 3.00656 & 4.99782 & 6.99826 \\
0.999559 & 3.00183 & 5.00132 & 6.99979 \\
0.99951 & 2.9991 & 5.00006 & 7.00021 \\
1.00011 & 2.99992 & 4.99986 & 7. \\
1.00004 & 3.0001 & 5.00001 & 6.99998 \\
0.999984 & 2.99999 & 5.00001 & 7. \\
0.999998 & 2.99999 & 5. & 7. \\
1. & 3. & 5. & 7. \\
1. & 3. & 5. & 7. \\
\end{array}
\right)$$
Compare that to the exact result above.
Even though it is not diagonally dominant, the iterations still converged. For the reasons why, see: Gauss-Seidel Iterative Method and Necessary condition for Gauss–Seidel method to converge
Update If we use the inital starting point as $(0,0,0,0)^T$, we get
$$\left(
\begin{array}{cccc}
0 & 0 & 0 & 0. \\
3.5 & 9.8 & 9.61429 & 6.73265 \\
-0.857823 & -0.404762 & 4.93178 & 7.68789 \\
1.46418 & 2.95145 & 4.53367 & 6.94695 \\
1.09466 & 3.33888 & 5.06521 & 6.93533 \\
0.943431 & 2.96412 & 5.03991 & 7.01357 \\
0.997067 & 2.96945 & 4.9883 & 7.00497 \\
1.00621 & 3.00752 & 4.99735 & 6.99798 \\
0.999525 & 3.00221 & 5.00154 & 6.99974 \\
0.999419 & 2.99895 & 5.00008 & 7.00024 \\
1.00012 & 2.9999 & 4.99983 & 7. \\
1.00004 & 3.00012 & 5.00001 & 6.99998 \\
0.999982 & 2.99999 & 5.00002 & 7. \\
0.999998 & 2.99999 & 5. & 7. \\
1. & 3. & 5. & 7. \\
1. & 3. & 5. & 7. \\
\end{array}
\right)$$
It converges just fine.
Best Answer
The Gauss-Seidel Method requires the matrix to be in diagonally dominant form.
This matrix is not diagonally dominant and G-S does not converge (sometimes it still may).
The first step is to put the matrix in D-D form so we have $Ax = b$ as
$$A = \begin{pmatrix}200 & -3 & 2 \\ 1 & -500 & 2 \\ 1 & -3 & 100 \\ \end{pmatrix}, b = \begin{pmatrix}765\\ 987 \\ 123 \\ \end{pmatrix}$$
You can figure out how many iterations you need to meet your accuracy requirement
$$\begin{align}x_0 &= (3.5,-2,1)\\ x_1 &= (3.785,-1.96243,1.1332771)\\ x_2 &= (3.784230779,-1.96189843,1.133300739)\\ x_3 &= (3.784238516,-1.96189832,1.133300665)\\ x_4 &= (3.784238519,-1.96189832,1.133300665)\end{align}$$
The exact results are
$$x = \begin{pmatrix} \dfrac{900785}{238036} \\ -\dfrac{3269017}{1666252} \\ \dfrac{3776729}{3332504} \\ \end{pmatrix} \approx \begin{pmatrix} 3.784238518543414\\-1.961898320302091\\1.133300665205503 \end{pmatrix}$$