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For $GL(3)$, the exterior square $L$-function $L(s, \pi, \wedge^2 \pi)$ is entire as it agrees with $L(s,\tilde\pi\otimes\omega)$, where $\omega$ is the central character of $\pi$. Therefore, $L(1,\pi,\mathrm{sym}^2)=0$ would imply that $L(1,\pi\otimes\pi)=0$, which is absurd.