homework and exercises - Spontaneous symmetry breaking by two scalar multiplets

Consider a theory with two multiplets of real scalar fields $\phi_i$ and $\epsilon_i$, where $i$ runs from $1$ to $N$. The Lagrangian is given by: $$\mathcal L = \frac{1}{2} (\partial_{\mu} \phi_i) (\partial^{\mu} \phi_i) + \frac{1}{2} (\partial_{\mu} \epsilon_i) (\partial^{\mu} \epsilon_i) − \frac{m^2}{2}[\phi_i \phi_i+ \epsilon_i \epsilon_i] − \frac{g}{8}[(\phi_i \phi_i)(\phi_j \phi_j ) + (\epsilon_i \epsilon_i)(\epsilon_j \epsilon_j)] − \frac{λ}{2}(\phi_i \epsilon_i)(\phi_j \epsilon_j ),$$ where $m^2 < 0, g > 0$ and $\lambda > −g/2.$ Summation over repeated indices is implied.

Is the following accurate? The lagrangian can be written in vector notation and we can see it is then invariant under a simultaneous transformation of $\phi$ and $\epsilon$ such that $\epsilon \rightarrow R_{ij}\epsilon_j$ and $\phi_i \rightarrow R_{ij} \phi_j$ if $R_{ik} R_{ij} = \delta_{kj}$ The symmetry group is then $O(N) \otimes O(N)$ with generators $T_a^{O(N) \otimes O(N)} = T_a^{O(N)} \otimes \text{Id}_{N \times N} + \text{Id}_{N \times N} \otimes T_a^{O(N)}$ so there are $\text{dim}O(N)$ number of generators.

The vacua of the theory can be found as the minimum of the potential $$V(\phi, \epsilon) = \frac{m^2}{2} ( \vec \phi^T \vec \phi + \vec \epsilon^T \vec \epsilon) + \frac{g}{8} ((\vec \phi^T \vec \phi)^2 + (\vec \epsilon^T \epsilon)^2) + \frac{\lambda}{2} (\vec \phi^T \vec \epsilon)^2$$ I am a bit confused here - to find the vacua I could demand $$\frac{\partial V}{\partial \phi^T \phi} = \frac{\partial V}{\partial \epsilon^T \epsilon} \overset{!}{=} 0$$ but what happens to the term proportional to $\lambda$?