JEE Math Practice Question

Let A = $$\left[ {\begin{matrix} 2 & b & 1 \\ b & {{b^2} + 1} & b \\ 1 & b & 2 \\ \end{matrix} } \right]$$ where b > 0. Then the minimum value of $${{\det \left( A \right)} \over b}$$ is -

1. A.$$\sqrt 3 $$
2. B.$$-$$ $$2\sqrt 3 $$
3. C.$$ - \sqrt 3 $$
4. D.$$2\sqrt 3 $$