$\int_{\partial\Omega_t}(h-H)|\frac{\partial u}{\partial \nu}|^2=0 \Rightarrow$ $\Omega_t$ is ball

Question

Consider the firt eigenvalue $\lambda$ of Laplacian on bounded mutative domain $\Omega_t\subset \mathbb R^n$ with smooth boundary $\partial \Omega_t$. By my calculation, $u$ is the eigenfunction of $\lambda$, I have $$ \frac{d\lambda}{dt} =\int_{\partial\Omega_t} \frac{\partial u}{\partial t }\nabla u\cdot \nu $$ If we deform the boundary $\partial\Omega_t$ by volume preserving mean curvature flow $$ \partial_tx=(h(t)-H(x,t))\nu ~~~x\in\partial \Omega_t $$ $\nu$ is the outer normal vector. $H(x,t)$ is the mean curvature , and $h(t)$ is the mean of $H(x,t)$ i.e. $$ h(t)=\frac{\int_{\partial\Omega_t} H dS}{\int_{\partial\Omega_t} dS} $$ then, I have $$ \frac{d\lambda}{dt}=\int_{\partial\Omega_t}(h-H)|\frac{\partial u}{\partial \nu}|^2 $$ I feel it is hard to know $\frac{d\lambda}{dt}$ is negative or positive, I have asked here. So, step back, I want to show it is ball when $$ \int_{\partial\Omega_t}(h-H)|\frac{\partial u}{\partial \nu}|^2=0 $$ But I still don't know how to do it.