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, $\frac{f\left(x\right)-14}{{\left(x-1\right)}^2}=0\left(x\ne 1\right)$ is: