Answer: $B$
Given: $$\ln \underbrace{(3x^2-4x+5)}_\text{f(x)} $$
Let $y =\ln \underbrace{(3x^2-4x+5)}_\text{f(x)} $
Let's suppose $f(x) = 3x^2 - 4x + 5$
Here, $a = 3, \;b = -4,\;and\;c = 5$
$Discriminant, D = b^2-4ac = -4^2-4.3.5 = 16-60 = -44 < 0$
Since, $D$ is negative, $\therefore$ $x$ can take any value in $\mathbb{R}$ which means it has no real roots $\implies$ For any value of $x$, $f(x)$ will always be $\bf{positive}$.
In $\ln f(x)$, $f(x)$ should always be "greater than 0" but as we have seen above $f(x)$ is always greater than 0 for any value of x.
$\therefore$ the Domain of $y$ is a set of all Real numbers, $\bf{\mathbb{R}}$
Hence, option $B$ is the correct answer.