Let $f(n)=\Omega(n), g(n)=O(n)$ and $h(n)=\theta(n)$ then $f(n).g(n)+h(n)$ is?
You have a fault in $f(n).g(n) = \Omega(n^2)$. Because, $g(n) = O(n)$ and you can't say $g(n)$ is $n$ or $\frac{1}{n^2}$ for example. Hence, we can't say nothing about $f(n).g(n)$. For more example:
$$g(n) = \frac{1}{n}, f(n) = n \Rightarrow f(n).g(n) = \Theta(1)$$
$$g(n) = 1, f(n) = n \Rightarrow f(n).g(n) = \Theta(n)$$
$$g(n) = \frac{1}{n^2}, f(n) = n \Rightarrow f(n).g(n) = o(1) (\text{little-o})$$ 
$$g(n) = n, f(n) = n \Rightarrow f(n).g(n) = \Theta(n^2)$$ 

As we don't know about $f(n).g(n)$. However, we can say $f(n).g(n) + h(n) = \Omega(n)$ as $h(n) = \Theta(n)$ and $f(n).g(n) \geq 0$.