The sum of all real roots of equation $$\left(e^{2x} - 4\right)\left(6e^{2x} - 5e^x + 1\right) = 0$$ is

We need to find the sum of all real roots of $$\left(e^{2x} - 4\right)\left(6e^{2x} - 5e^x + 1\right) = 0$$.

Since $$e^{2x} - 4 = 0$$, it follows that

$$ e^{2x} = 4 \implies e^x = 2 \implies x = \ln 2 $$

Next, setting the second factor to zero gives $$6e^{2x} - 5e^x + 1 = 0$$, and letting $$t = e^x$$ (where $$t > 0$$) leads to

$$ 6t^2 - 5t + 1 = 0 $$

Using the quadratic formula, we obtain

$$ t = \frac{5 \pm \sqrt{25 - 24}}{12} = \frac{5 \pm 1}{12} $$

This yields

$$ t = \frac{6}{12} = \frac{1}{2} \quad \text{or} \quad t = \frac{4}{12} = \frac{1}{3} $$

Substituting back, for $$t = \frac{1}{2}$$ we have $$e^x = \frac{1}{2} \implies x = -\ln 2$$, and for $$t = \frac{1}{3}$$ we have $$e^x = \frac{1}{3} \implies x = -\ln 3$$.

Thus, the three real roots are $$\ln 2, -\ln 2, -\ln 3$$, and their sum is

$$ \ln 2 + (-\ln 2) + (-\ln 3) = -\ln 3 $$

Therefore, the correct option is Option B: $$-\ln 3$$.