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 $L(1,\pi\otimes\pi)=0$, contradicting a result of Shahidi (1980). The best known zero-free region for general Rankin-Selberg $L$-functions is due to Brumley (2012): see the Appendix here.

For $GL(3)$, the exterior square $L$-function $L(s,\wedge^2\pi)$ is entire as it agrees with $L(s,\tilde\pi\otimes\omega)$, where $\omega$ is the central character of $\pi$. Therefore, $L(1,\mathrm{sym}^2\pi)=0$ would imply that $L(1,\pi\otimes\pi)=0$, contradicting a result of Shahidi (1980). The best known zero-free region for general Rankin-Selberg $L$-functions is due to Brumley (2012): see the Appendix here.

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.

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 $L(1,\pi\otimes\pi)=0$, contradicting a result of Shahidi (1980). The best known zero-free region for general Rankin-Selberg $L$-functions is due to Brumley (2012): see the Appendix here.