Radicals and Plotkin's problem concerning geometrically equivalent groups

If $G$ and $X$ are groups and $N$ is a normal subgroup of $X$, then the $G$-closure of $N$ in $X$ is the normal subgroup ${\overline X}^G = \bigcap \{ \ker \varphi \vert \varphi : X\rightarrow G, \mbox{ with } N \subseteq \ker \varphi \}$ of $X$. In particular, ${\overline 1}^G = R_GX$ is the $G$-radical of $X$. Plotkin calls two groups $G$ and $H$geometrically equivalent, written $G\sim H$, if for any free group $F$ of finite rank and any normal subgroup $N$ of $F$ the $G$-closure and the $H$-closure of $N$ in $F$ are the same. Quasi-identities are formulas of the form $(\bigwedge_{i\le n} w_i = 1 \rightarrow w =1)$ for any words $w, w_i (i\le n)$ in a free group. Generally geometrically equivalent groups satisfy the same quasi-identities. Plotkin showed that nilpotent groups $G$ and $H$ satisfy the same quasi-identities if and only if $G$ and $H$ are geometrically equivalent. Hence he conjectured that this might hold for any pair of groups. We provide a counterexample.