Vertices of an isosceles are: A(2, 0) and B(2, 5).

Let the third vertex is P(x, y)

On squaring both sides, we get

(x - 2)^{2} + (y)^{2} = (x - 2)^{2} + (y - 5)^{2}

x^{2} - 4x + 4 + y^{2} = x^{2} - 4x + 4 + y^{2} - 10y + 25

y = \(\frac{5}2,\)

Also, PA = 3

\(\sqrt{(x - 2)^2 + (y)^2} = 3\)

On squaring both sides, we get

(x - 2)^{2} + (y)^{2} = 9

x^{2} - 4x + 4 + y^{2} = 9

x^{2} - 4x + y^{2} = 5

On substituting \(y = \frac{5}2,\)

\(x^2 - 4x + \frac{25}4 = 5\)

\(x^2 - 4x + \frac{25} 4 - 5 = 0\)

\(x^2 - 4x + \frac{5}4 = 0\)

Using quadratic formula:

Therefore coordinates of third vertex are: