Given:

$f\left(x\right)=(x-1)(x-2)(x-3)$

$\Rightarrow f\left(x\right)=\left(x^3-6x^2+11x-6\right)$

$f^{\text{'}}\left(x\right)=3x^2-12x+11$

Now, $f\left(x\right)$ is continuous in $R$ as it is a odd degree polynomial.

From Lagrange's mean value theorem, if $f\left(x\right)$ is continuous in $\left[a,b\right]$ and differentiable in $\left(a,b\right)$ then there exist  $c\in \left(a,b\right)$ such that 

$f^{\text{'}}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

$\Rightarrow 3c^2-12c+11=\frac{f\left(4\right)-f\left(0\right)}{4-0}$

$\Rightarrow 3c^2-12c+11=\frac{6-\left(-6\right)}{4-0}$

$\Rightarrow 3c^2-12c+11=3$

$\Rightarrow 3c^2-12c+8=0$

$\Rightarrow c=\frac{12\pm \sqrt{144-12\times 8}}{6}$

$\Rightarrow c=\frac{12\pm \sqrt{48}}{6}$

$\Rightarrow c=\left(2\pm \frac{2\sqrt{3}}{2}\right)\in \left(0,4\right)$.