1. The midpoint M of BC is:
\[M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{1 - 2}{2}, \frac{3 + 7}{2}\right) = \left(-\frac{1}{2}, 5\right).\]
This midpoint serves as the point of symmetry for the isosceles triangle.
2. The slope of BC is:
\[m_{BC} = \frac{7 - 3}{-2 - 1} = -\frac{4}{3}.\]
3. The slope of the perpendicular bisector is the negative reciprocal of the slope of BC:
\[m_\text{perp} = \frac{3}{4}.\]
4. The equation of the perpendicular bisector passing through \(M\left(-\frac{1}{2}, 5\right)\) is:
\[y - 5 = \frac{3}{4}\left(x + \frac{1}{2}\right).\]
Simplify:
\[y - 5 = \frac{3}{4}x + \frac{3}{8} \Rightarrow y = \frac{3}{4}x + \frac{43}{8}.\]
5. The vertex A lies on this perpendicular bisector, and its distance from both B and C must be equal.
6. Using the distance formula between \(A(x,y)\) and \(B(1,3)\)
\[d_{AB} = \sqrt{(x-1)^2 + (y-3)^2}.\]
Similarly, the distance \(d_{AC}\) between \(A(x,y)\) and \(C(-2,7)\) is:
\[d_{AC} = \sqrt{(x+2)^2 + (y-7)^2}.\]
7. Equate \(d_{AB} = d_{AC}\):
\[\sqrt{(x-1)^2 + (y-3)^2} = \sqrt{(x+2)^2 + (y-7)^2}.\]
8. Square both sides and simplify:
\[(x-1)^2 + (y-3)^2 = (x+2)^2 + (y-7)^2.\]
Expand:
\[x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 + 4x + 4 + y^2 - 14y + 49.\]
Simplify:
\[-6x + 10 - 8y + 40 = 0 \Rightarrow -6x + 8y + 10 = 0 \Rightarrow -6x + 8y - 25 = 0.\]
9. Solve this equation along with the perpendicular bisector equation to find \(A(x,y)\). After solving, the possible coordinates of A are: \(\left(-\frac{5}{6}, 6\right)\), \((-7, -\frac{1}{8})\)
Thus, the correct answers are (C) and (D).