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