Real roots are demanded!

First we need to find the local maximum and minimum values of $f(x)$ concerned.

$\dfrac {d}{dx} f(x) = f'(x) = 6x^2-18x+12$

$f'(x) = 0 \quad \Rightarrow x^2-3x+2 = (x-1)(x-2)=0\quad \Rightarrow x = 1 \text{ or } 2$

$\begin{cases} f(1) = 2(1)-9(1)+12(1)-k = 5-k \\ f(2) = 2(8)-9(4)+12(2)-k = 4-k \end{cases}$

For all the three roots of $f(x)$ to be real,

$\begin{cases} \text{The maximum: } & 5-k > 0 & \Rightarrow k < 5 = n \\ \text{The minimum: } & 4-k < 0 & \Rightarrow k > 4 = m \end{cases}$

$\Rightarrow m\times n \times (n-m) = 4\times 5 \times (5-4) = \boxed{20}$

For a cubic equation $ax^3+bx^2+cx+d=0$ , its discriminant must be greater than $0$ to have three distinct real roots, and it is:

$\Delta>0 \\ b^2c^2-4ac^3-4b^3d+18abcd-27a^2d^2>0 \\ (-9)^2(12)^2-4(2)(12)^3-4(-9)^3(-k)+18(2)(-9)(12)(-k)-27(2)^2(-k)^2>0 \\ -108k^2+972k-2160>0 \\ 108(-k^2+9k-20)>0 \\ -k^2+9k-20>0 \\ -(k-4)(k-5)>0 \\ 4<k<5$

Hence, $m=4$ , $n=5$ and $mm(n-m)=\boxed{20}$ .

$f(x)=2x^3-9x^2+12x-k$

$f'(x)=6x^2-18x+12$ and at $f('x)=0$ , x=1,2. With $f('x)$ having 2 real roots, $f(x)$ has a minima and a maxima. For the function to have have real roots: the maxima should be above the x-axis and the minima below the x-axis, or the minima or the maxima lies on the x-axis. When either the minima or the maxima lies on the x-axis, it must be a repeated root.

Solving for k when $f(2)=0$ , $k=4$ ; $f(1)=0$ , $k=5$ . Thus, $m=4$ and $n=5$ . So $m\times n\times(n-m)=20$

Draw graph for 2x^3 - 9x^2 + 12x. You will observe that it intersects at 0 only. Local maximum at 5 and minimum at 4 is found to appear. If you decrease the 'y' value of this function by 5 you will get 2 repeating roots and 1 distinct root. If you decrease the 'y' value of this function by 4 you will get 2 repeating roots and 1 distinct again. But here the repeating roots will be greater than the distinct root. These are the critical cases. Hence k belongs to (4, 5) .

Excuse my non latex answer!!