Let $f\left(x\right)=x^3-3\left(a-2\right)x^2+3ax+7$ and $f\left(x\right)$ is increasing in $\left(0,1\right]$ and decreasing is $\left[1,5\right)$, then roots of the equation $\frac{f\left(x\right)-14}{{\left(x-1\right)}^2}=0$ is (A) $1$ (B) $3$ (C) $7$ (D) $-2$

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