Let $\Omega \subset \mathbb{R}^3$ a smooth bounded domain with $0\in \Omega$ and $u_\varepsilon(x)$ the solution to 
$$
\Delta u_\varepsilon + (\lambda-\varepsilon) u_\varepsilon = \frac{1}{|x|}\quad \text{in } \Omega,
\quad u_\varepsilon(x)=0 \quad \text{on }\partial \Omega.$$
Here $\lambda_{k}<\lambda-\varepsilon<\lambda<\lambda_{k+1}$ for some $\lambda_k \in \sigma(-\Delta)$.
I am interested to *find sufficient conditions on the location of $0\in \Omega$* to have $u_\varepsilon(0)\neq 0$ for almost every $\varepsilon \in [0,\varepsilon_1)$.

For example, I would like to *prove that locating the origin near the boundary is sufficient*: if $0<d(0,\partial \Omega)<\varepsilon_0$ for some $\varepsilon_0$, then $u_\varepsilon(0)\neq 0$ for almost every $\varepsilon \in [0,\varepsilon_1)$ for some $\varepsilon_1=\varepsilon_1(\varepsilon_0)$.

Some observation:
 1. I worked on the problem for $v_\varepsilon\in C^{2,1}(\Omega)$ where $u_\epsilon(x)=\frac{1}{2}|x|+v_\varepsilon(x)$. The problem becomes
$$
\Delta v_\varepsilon + (\lambda - \varepsilon)v_\varepsilon =-\lambda\frac{|x|}{2}\quad \text{in } \Omega,\quad v_\varepsilon(x)=-\frac{|x|}{2}\quad \text{on } \partial \Omega
$$
 2. For $d(0,\partial \Omega)=|x|=\delta(\varepsilon)$ small I thought it could be useful to expand $-\frac{1}{2}\delta=\sum_{k=0}^\infty D_x^k \theta(0)\cdot x^k$ and using the strong continuation property to get a contradiction, but I don't see how to get it.
 3. Maximum principle does not apply.

 Any reference or suggestion on would be helpful.