Let $G$ be a linear algebraic group -- be it a Lie group or a group of Lie type. Let $V$, $W$ be subvarieties of $G$. Of course, $V\cap W$ is also a variety (not necessarily irreducible) and $V\cdot W^{-1}$ is a constructible set. It is easy to see that $\dim(\overline{V\cdot W^{-1}}) = \dim(V)+\dim(W)-\dim(V\cap a W)$ for $a\in V\cap W$ generic. By Bézout, we know that the degree $\deg(V\cap a W)$ of $V\cap a W$ (meaning the sum of the degrees of its components) is at most $\deg(V)\cdot \deg(W)$.

Is it always the case that $\deg(V \cap a W) \cdot \deg(\overline{V\cdot W^{-1}}) \leq \deg(V)\cdot \deg(W)$, for $a\in \overline{V\cdot W^{-1}}$ generic?

If not, is that the case under some sort of generic conditions, or, say, for $W = g V g^{-1}$ and $g\in G$ generic?

Reason why such a thing might be plausible: if $\dim(V) + \dim(W)\geq \dim(G)$, then, generically, $\overline{V\cdot W}$ is the entire group, and, while $\deg(G)$ is not in general $1$ for $G$ seen as a subvariety of $M(n)\sim \mathbb{A}^{n^2}$, the degree of $G$ "relative" to itself is in some sense $1$ (of course we would need to formalise that notion; the point is that that's how it works out in further applications of Bézout's theorem, in a trivial sense, in so far as, for $V\subset G$, $\deg(V\cap G) = \deg(V) \cdot 1$). If $\dim(V)+\dim(W)<\dim(G)$, then, generically, the intersection $V\cap W$ is empty, and non-generically, Bézout may not be tight.