Given, $(2 a)^{\ln a}=(b c)^{\ln b}$, where $2 a>0, b c>0$

$$ \Rightarrow \ln a(\ln 2+\ln a)=\ln b(\ln b+\ln c) $$ ..........(i)

and $(b)^{\ln 2}=(a)^{\ln c}$

$$ \Rightarrow \ln 2 \cdot \ln b=\ln c \cdot \ln a $$ ..........(ii)

Now, let $\ln a=x, \ln b=y$

$$ \ln 2=p, \ln c=z $$

Now, from Eqs. (i) and (ii), we get

$$ p \cdot y=x z \text { and } x(p+x)=y(y+z) $$

$$ \begin{aligned} & \therefore p=\frac{x z}{y} \\\\ & \therefore x\left(\frac{x z}{y}+x\right)=y(y+z) \\\\ & \Rightarrow x^2 z+x^2 y=y^2(y+z) \\\\ & \Rightarrow \left(x^2-y^2\right)(y+z)=0 \end{aligned} $$

$$ \therefore $$ $ x^2 = y^2 $

$ x = \pm y $

So, from the equations, there are two cases :

Case 1 :

$ x = y $

In this case, since x and y are natural logarithms of positive numbers a and b respectively, this implies that a = b. However, this cannot be true as a, b, and c are given to be distinct positive real numbers.

Case 2 :

x = -y

In this case, $ \ln a = -\ln b $

$ \Rightarrow a \times b = 1$

$ \Rightarrow b = \frac{1}{a} $

Also, y + z = 0 (from the equations above)

$ \Rightarrow \ln b + \ln c = 0 $

$ \Rightarrow \ln (b \times c) = 0 $

$ \Rightarrow b \times c = 1 $

Given $ b = \frac{1}{a} $ and $ b \times c = 1 $ $ \Rightarrow c = a $

Thus, in the case where x = -y, the possible values are :

$b = \frac{1}{a} $

$c = a$

$$ \begin{aligned} & \text { If } b c=1 \Rightarrow(2 a)^{\ln a}=1 \\\\ & \Rightarrow a=1 / 2 \\\\ & \text { So, } 6 a+5 b c=6\left(\frac{1}{2}\right)+5=3+5=8 \end{aligned} $$