Let a ,   b ,   c be the three distinct positive real numbers such that 2 a log e a = b c log e b…

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

$\Rightarrow \ln \left(a\right)·\left(\ln \left(2a\right)\right)=\ln \left(b\right)·\ln \left(bc\right)$

$\Rightarrow \ln \left(a\right)·\left(\ln \left(2\right)+\ln \left(a\right)\right)=\ln \left(b\right)·\left(\ln \left(b\right)+\ln \left(c\right)\right)$

let $\ln \left(a\right)=x, \ln \left(b\right)=y, \ln \left(c\right)=z, x\ne y\ne z$

$\Rightarrow x\left(\ln \left(2\right)+x\right)=y\left(y+z\right)$

$\Rightarrow x\ln \left(2\right)=y^2-x^2+yz$ ........(i)

Similarly, from second equation

$\ln \left(2\right)·\ln \left(b\right)=\ln \left(c\right)·\ln \left(a\right)$

$\Rightarrow \ln \left(2\right)=\frac{xz}{y}$ ......(ii)

Substitute $eq\left(ii\right) \text{in} eq\left(i\right)$,

We get,

$\Rightarrow y^3-y{x}^2+y^{2}z=x^{2}z$

$\Rightarrow y^2\left(y+z\right)-x^2\left(y+z\right)=0$

$\Rightarrow \left(y^2-x^2\right)\left(y+z\right)=0$

$\Rightarrow \left(x-y\right)\left(x+y\right)\left(y+z\right)=0$

$\because x\ne y\Rightarrow \left(x+y\right)\left(y+z\right)=0$

Now $x=-y, y=-z \text{and} x=z$

But $\ln \left(2\right)=\frac{xz}{y}$

$\Rightarrow \ln \left(2\right)=\frac{x\left(x\right)}{-x}$

$\Rightarrow \ln \left(2\right)=-\ln \left(a\right)$

$=-\ln 2\Rightarrow a=\frac{1}{2}$ and

$\Rightarrow y=-z$

$\Rightarrow \ln \left(b\right)=-\ln \left(c\right)$

$\Rightarrow bc=1$

Now $6a+5bc=6\left(\frac{1}{2}\right)+5\left(1\right)=8$

Hence this is the required answer.