Explanation:

Let the centre be \((h, k)\) which will be equidistant from all 3 points.

\(Radius^2 = (11 - h)^2 + (0 - k)^2 = (7 - h)^2 + (0 - k)^2 = (9 - h)^2 + (2\sqrt{3} - k)^2\)

Using \((11 - h)^2 + (0 - k)^2 = (7 - h)^2 + (0 - k)^2\)
\(121 + h^2 - 22h + k^2= 49 + h^2 - 14h + k^2\)
\(121 - 49 = 8h\)
\(72 = 8h\)
\(h = 9\)

Plugging in the value of \(h = 9\) in \((7 - h)^2 + (0 - k)^2 = (9 - h)^2 + (2\sqrt{3} - k)^2\);

\((7 - 9)^2 + (0 - k)^2 = (9 - 9)^2 + (2\sqrt{3} - k)^2\)
\(4 + k^2 = 0 + 12 + k^2 - 4\sqrt{3}k\)
\(4\sqrt{3}k = 8\)
\(\sqrt{3}k = 2\)
\(k = \frac{2}{\sqrt{3}}\)

So, the centre is \((9, \frac{2}{\sqrt{3}})\)

\(Radius^2 = (9 - 7)^2 + (\frac{2}{\sqrt{3}} - 0)^2 = 4 + \frac{4}{3} = \frac{16}{3}\)

\(Radius = \frac{4}{\sqrt{3}}\)
Diameter = \(2(\frac{4}{\sqrt{3}}) = \frac{8}{\sqrt{3}}\)

Col. A: \(\frac{8}{\sqrt{3}}\)
Col. B: \(4\)

Dividing by \(4\) and multiplying by \(\sqrt{3}\) on both sides;

Col. A: \(2\)
Col. B: \(\sqrt{3}\)

Hence, option A