设函数\(f\left( x \right)=\begin{cases} & \left| 12x-4 \right|,x\leqslant 1 \\ & x{{\left( x-2 \right)}^{2}}+a,x > 1 \end{cases}\)，若存在互不相等的\(4\)个实数\({{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}\)，使得\(\dfrac{f\left( {{x}_{1}} \right)}{{{x}_{1}}}=\dfrac{f\left( {{x}_{2}} \right)}{{{x}_{2}}}=\dfrac{f\left( {{x}_{3}} \right)}{{{x}_{3}}}=\dfrac{f\left( {{x}_{4}} \right)}{{{x}_{4}}}=7\)，则\(a\)的取值范围为(    )