Does the symmetric square L-function vanish at one?

Take a cuspidal automorphic representation $$\pi$$ of $$GL(3)$$ over a number field. My question is quite straightforward and can be related to this one :

Can $$L(1, \pi, \mathrm{sym}^2)$$ be zero? If yes, is there any extra assumption ensuring it cannot?

I know that we have the splitting $$L(s, \pi \times \tilde{\pi}) = L(s, \pi, \mathrm{sym}^2)L(s, \pi, \wedge^2 \pi)$$. If we assume $$\pi$$ self-contragredient for instance, then the Rankin-Selberg convolution has a simple pole at $$s=1$$. Either $$L(s, \pi, \mathrm{sym}^2)$$ has a pole at $$s=1$$ (Gelbart-Jacquet lift case) and in that case it does not vanish ; or it has none and in that case the question rephrases as : can $$L(s, \pi, \wedge^2 \pi)$$ have a pole of order 2 at $$s=1$$?

Any new insight or reference on this precise question is welcome.

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.