The correct answer is $\boxed{\text{D}}$.
To find the limit, we can substitute $x=3$ into the function. However, this results in the indeterminate form $\frac{0}{0}$. This means that the two expressions have the same value, but we cannot determine what that value is by simply substituting.
To resolve this, we can use L’Hôpital’s rule. L’Hôpital’s rule states that if $$\lim_{x\to a} \frac{f(x)}{g(x)} = \frac{0}{0}$$ or $$\lim_{x\to a} \frac{f(x)}{g(x)} = \pm\infty$$, then $$\lim_{x\to a} \frac{f'(x)}{g'(x)}$$ exists and is equal to the limit of the original expression.
In this case, we have $$\begin{align}
\lim_{x\to 3} {\text{f}}\left( {\text{x}} \right) &= \lim_{x\to 3} \frac{{2{{\text{x}}^2} – 7{\text{x}} + 3}}{{5{{\text{x}}^2} – 12{\text{x}} – 9}} \
&= \lim_{x\to 3} \frac{{2\left( {\text{x}}^2 – \frac{7}{2}{\text{x}} + \frac{3}{2}} \right)}}{{5\left( {\text{x}}^2 – \frac{12}{5}{\text{x}} – \frac{9}{5}} \right)}} \
&= \lim_{x\to 3} \frac{{2\left( {\text{x} – \frac{3}{2}} \right)}^2}}{{5\left( {\text{x} – \frac{3}{5}} \right)}^2}} \
&= \frac{{2\left( 3 – \frac{3}{2}} \right)}^2}}{{5\left( 3 – \frac{3}{5}} \right)}^2}} \
&= \frac{5}{18}
\end{align}$$ Therefore, $$\mathop {\lim }\limits_{{\text{x}} \to 3} {\text{f}}\left( {\text{x}} \right) = \frac{5}{{18}}.$$
The other options are incorrect because they do not represent the value of the limit.