Answer
$$ L(t)= \langle-3,10,16 \rangle +t \langle-4,5,24 \rangle .$$
Work Step by Step
Since $ r (t)=\langle1-t^2,5t,2t^3 \rangle, $ then $ r' (t)=\langle-2t,5,6t^2 \rangle $
then, the parametrization of the tangent line at $ t=2$ is given by
$$ L(t)=r (2)+tr'(2)=\langle-3,10,16 \rangle +t \langle-4,5,24 \rangle .$$
