Let $a,b,c$ be three distinct positive real numbers $\left(ab\ne 1\right)$ such that ${\left(2a\right)}^{{\log }_{e}a}={\left(bc\right)}^{{\log }_{e}b}\mathrm{and}{b}^{{\log }_{e}2}=a^{{\log }_{e}c}$  Then $6a+6bc$ is equal to

Given equation  ${\left(2a\right)}^{\ln a}={\left(bc\right)}^{\ln b}$

$$\begin{gathered} \ln a\times \ln \left(2a\right)=\ln \left(b\right).\ln \left(bc\right) \\ \Rightarrow \ln a(\ln 2+\ln a)=\ln b.(\ln b+\ln c) \\ \ln a=x,\ln b=y,\ln c=z,x\ne y\ne z \\ x(\ln 2+x)=y(y+z) \\ x\ln 2=y^2-x^2+yz \end{gathered}$$

Similarly from second equation
 $$\begin{gathered} \ln 2.\ln b=\ln c.\ln a \\ \ln 2=\frac{xz}{y} \end{gathered}$$
Substitute eq(ii) and eq(i)
We get,
$$\begin{gathered} \Rightarrow y^3-y{x}^2+y^{2}z=x^{2}z \\ {y}^2(y+2)-x^2(y+z)=0 \\ (x-y)(x+y)(y+z)=0 \\ x\ne y\Rightarrow (x+y)(y+z)=0 \\ \Rightarrow y=-z \end{gathered}$$

But  $\ln 2=\frac{xz}{y}$

$$\begin{gathered} \Rightarrow \ln 2=\frac{xz}{(-z)} \\ \ln 2=-x=-\ln a \\ a=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \Rightarrow y=-z \\ \ln b=-\ln c \\ \ln b+\ln c=0 \\ bc=1 \\ 6a+6bc=6\left(\frac{1}{2}\right)+6\left(1\right)=9 \end{gathered}$$