\nonumber   \\
&& + \, [\sigma(\vec{e}\,^{\prime}_{1})\otimes \sigma(\vec{e}\,^{\prime}_{2}) -
\sigma(\vec{e}_1)\otimes \sigma(\vec{e}_{2})]\otimes
\sigma(\vec{e}_{3}) .\end{aligned}$$ As in the two-particle case, it can be shown [@Cereceda] that in order for the three-particle state (2) to yield the largest violation of the inequality (20) it is necessary that $|c_1|=|c_2|=1/\sqrt{2}$. However, the largest eigenvalue of the Bell operator (21) is given by [@Cereceda] $$\begin{aligned}
\lambda_l = && 2 \, (1+|\sin\theta_1 \sin\theta_2| + |\sin\theta_2 \sin\theta_3| \nonumber \\
&& + \, |\sin\theta_1 \sin\theta_3|)^{1/2} ,\end{aligned}$$ so that a maximum violation occurs provided that $\vec{e}_i \perp \vec{e}\,^{\prime}_{i}$ for each $i=1,2,3$. The corresponding eigenvector will consist of an equally weighted superposition of the states $|\sigma_{\perp 1}\rangle|\sigma_{\perp 2}\rangle|\sigma_{\perp 3}\rangle$ and $|-\sigma_{\perp 1}\rangle|-\sigma_{\perp 2}\rangle|-\sigma_{\perp 3}\rangle$ with a relative phase between them. As before, $\theta_i$ is the angle between the vectors $\vec{e}_i$ and $\vec{e}\,^{\prime}_{i}$, and $|\sigma_{\perp i}\rangle$ ($|-\sigma_{\perp i}\rangle$) denotes the eigenvector \[with eigenvalue $\sigma_{\perp i}=\pm 1$ ($-\sigma_{\perp i}=\mp 1$)\] of the spin operator for particle $i$ along the direction perpendicular to both $\vec{e}_i$ and $\vec{e}\,^{\prime}_{i}$. Likewise, the relative signs of the eigenvalues $\sigma_{\perp 1}$, $\sigma_{\perp 2}$, and $\sigma_{\perp 3}$ entering the superposition will depend on the relative signs of $\sin\theta_1$, $\sin\theta_2$, and $\sin\theta_3$. \[Note that the maximum amount of violation of inequality (20) 