To find the sum of all the real roots of the given equation, we need to first solve the equation and then add up all the real roots.

Given equation: (e^2x - 4)(6e^2x - 5ex + 1) = 0

To solve this equation, we will first find the roots of each factor separately and then combine them.

1. Finding the roots of the first factor (e^2x - 4):
Setting the first factor equal to zero, we have:
e^2x - 4 = 0

Adding 4 to both sides:
e^2x = 4

Taking the natural logarithm of both sides:
ln(e^2x) = ln(4)

Using the property of logarithms (ln(e^a) = a), we get:
2x = ln(4)

Dividing both sides by 2:
x = ln(4)/2

So, the first factor has one real root at x = ln(4)/2.

2. Finding the roots of the second factor (6e^2x - 5ex + 1):
This factor is a quadratic equation in terms of e^x. We can use the quadratic formula to find its roots.

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the roots are given by:
x = (-b ± √(b^2 - 4ac))/(2a)

For the given quadratic equation, a = 6, b = -5e, and c = 1.
Plugging these values into the quadratic formula, we get:
x = (-(-5e) ± √((-5e)^2 - 4*6*1))/(2*6)

Simplifying, we have:
x = (5e ± √(25e^2 - 24))/(12)

Since the discriminant (b^2 - 4ac) determines the nature of the roots, we need to determine whether it is positive, negative, or zero.

Calculating the discriminant:
D = (5e)^2 - 4*6*1
D = 25e^2 - 24

Since the value of e is approximately 2.71828 (Euler's number), we can substitute it in the discriminant to find its value.

D = 25(2.71828)^2 - 24
D ≈ 183.93833

Since the discriminant is positive, the quadratic equation has two real roots.

Therefore, the second factor has two real roots.

3. Summing up all the real roots:
The first factor has one real root, and the second factor has two real roots.

Summing up all the real roots:
Sum = ln(4)/2 + [(5e + √(25e^2 - 24))/(12)] + [(5e - √(25e^2 - 24))/(12)]

Simplifying further:
Sum = ln(4)/2 + (10e)/(12)

Using the approximation for e, we can calculate the value of the sum:
Sum ≈ 0.54931

The correct answer is option B) -loge3.