JEE Math Practice Question

If the function f given by f(x) = x3 – 3(a – 2)x2 + 3ax + 7, for some a$$ \in $$R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, $${{f\left( x \right) - 14} \over {{{\left( {x - 1} \right)}^2}}} = 0\left( {x \ne 1} \right)$$ is :

1. A.$$-$$ 7
2. B.5
3. C.7
4. D.6