Given that, $|3x-20| + |3x-40| = 20 ; x \in \mathbb{R} \quad \longrightarrow (1)$
We know that $,|x| = \left\{\begin{matrix} x\;;&x\geq 0 \\ -x\;; &x<0 \end{matrix}\right.$
We can open mod as positive and negative. There are four such cases.
$\textbf{Case 1:}\;\text{ Positive, Positive}$
$\Rightarrow 3x – 20 + 3x – 40 = 20$
$\Rightarrow 6x – 60 = 20$
$\Rightarrow 6x = 80$
$\Rightarrow \boxed{x = \frac{40}{3} = 13.33}$
$\textbf{Case 2:}\;\text{ Positive, Negative}$
$\Rightarrow 3x – 20 – (3x – 40) = 20$
$\Rightarrow 3x – 20 – 3x + 40 = 20$
$\Rightarrow \boxed{20 = 20\; {\color{Green} {\text{(True)}}}}$
$\textbf{Case 3:}\;\text{ Negative, Positive}$
$\Rightarrow \;– (3x – 20) + 3x – 40 = 20$
$\Rightarrow\; – 3x + 20 + 3x – 40 = 20$
$\Rightarrow \boxed{- 20 = 20\;\color{Red}{\text{(False)}}}$
$\textbf{Case 4:}\;\text{ Negative, Negative}$
$\Rightarrow \;– (3x – 20) – (3x – 40) = 20$
$\Rightarrow \;– 3x + 20 – 3x + 40 = 20$
$\Rightarrow \;– 6x =\; – 40$
$\Rightarrow \boxed{x = \frac{20}{3} = 6.66}$
$\therefore$ $\boxed{7 < x < 12}$
Correct Answer $: \text{C}$