At which points on the curve $\alpha(t):=(3t-t^3,3t^2,3t+t^3)$ the corresponding tangent lines are parallel to the plane $3x+y+z+2=0$?
This will give you a polynomial equation in $t$.
And we get
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$n=(3,1,2)$ is a normal vecttor to the plane . We have : $\alpha'(t)=(3-3t^2,6t,3+3t^2)$ is a tangent line director vector is parallel to the plane if it's normal to $n$, thus : $n.\alpha'(t)=0$, thus : $$3(1-t^2) + 2t + 1+t^2=0$$ $$-2t^2 + 2t +4 =0 $$ $$t^2-t-2 =0 $$ gives : $t=-1$ or $t=2$