Let a, b, c be the three distinct positive real numbers such that...

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

$⇒ \ln \left(a\right) \cdot \left(\ln \left(2a\right)\right) = \ln \left(b\right) \cdot \ln \left(bc\right)$

$⇒ \ln \left(a\right) \cdot \left(\ln \left(2\right) + \ln \left(a\right)\right) = \ln \left(b\right) \cdot \left(\ln \left(b\right) + \ln \left(c\right)\right)$

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

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

$⇒ xln\left(2\right) = {y}^{2} - {x}^{2} + yz$ ........(i)

Similarly, from second equation

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

$⇒ \ln \left(2\right) = \dfrac{xz}{y}$ ......(ii)

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

We get,

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

$⇒ {y}^{2}\left(y + z\right) - {x}^{2}\left(y + z\right) = 0$

$⇒ \left({y}^{2} - {x}^{2}\right)\left(y + z\right) = 0$

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

$∵ x \ne y ⇒ \left(x + y\right)\left(y + z\right) = 0$

Now $x = - y, y = - z\text{and}x = z$

But $\ln \left(2\right) = \dfrac{xz}{y}$

$⇒ \ln \left(2\right) = \dfrac{x\left(x\right)}{- x}$

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

$= - \ln 2 ⇒ a = \dfrac{1}{2}$ and

$⇒ y = - z$

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

$⇒ bc = 1$

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

Hence this is the required answer.