1. Given function

$f(x)=x^3-3(a-2)x^2+3ax+7$

We are told:

• $f$ is increasing on $(0,1]$
• $f$ is decreasing on $[1,5)$

We need a root of

$\frac{f(x)-14}{(x-1)^2}=0, \qquad x\ne 1$

Since the denominator is nonzero for $x\ne 1$, this equation is equivalent to

$f(x)-14=0, \qquad x\ne 1$

So we must solve

$f(x)=14$

2. Use monotonicity to determine $a$

First compute derivative:

$f'(x)=3x^2-6(a-2)x+3a$

Factor out $3$:

$f'(x)=3\left[x^2-2(a-2)x+a\right]$

Since $f$ is increasing on $(0,1]$ and decreasing on $[1,5)$, the point $x=1$ must be a turning point where the derivative changes from positive to negative. Hence,

$f'(1)=0$

Now,

$f'(1)=3-6(a-2)+3a=0$

$3-6a+12+3a=0$

$15-3a=0$

$a=5$

3. Substitute $a=5$ into $f(x)$

$f(x)=x^3-3(5-2)x^2+3(5)x+7$

$f(x)=x^3-9x^2+15x+7$

Now solve

$f(x)=14$

So,

$x^3-9x^2+15x+7=14$

$x^3-9x^2+15x-7=0$

4. Factor the cubic

Try $x=1$:

$1-9+15-7=0$

So $(x-1)$ is a factor.

Divide:

$x^3-9x^2+15x-7=(x-1)(x^2-8x+7)$

Further factor:

$x^2-8x+7=(x-1)(x-7)$

Hence,

$x^3-9x^2+15x-7=(x-1)^2(x-7)$

Thus,

$f(x)-14=(x-1)^2(x-7)$

So,

$\frac{f(x)-14}{(x-1)^2}=x-7, \qquad x\ne 1$

Therefore the root is

$x=7$

5. Check with options

The root is $7$.

So the correct option is:

C: $7$

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.