= Solution
Differentiation gives
$$
r'(t)=3\sin t\cos t
\left(-2\cos t,\sin t,\sqrt3\sin t\right),
$$
and $|r'|=6|\sin t\cos t|$. Thus
$$
T=\frac{\sigma}{2}
\left(-2\cos t,\sin t,\sqrt3\sin t\right),
\qquad
\sigma=\operatorname{sgn}(\sin t\cos t).
$$
The plus sign applies on $(0,\pi/2)\cup(\pi,3\pi/2)$ and the minus sign on $(\pi/2,\pi)\cup(3\pi/2,2\pi)$.
On each open quadrant,
$$
\boxed{
N=\sigma
\left(\sin t,\frac12\cos t,\frac{\sqrt3}{2}\cos t\right)},
\qquad
\boxed{
B=\left(0,\frac{\sqrt3}{2},-\frac12\right)}.
$$
The constant binormal proves that the curve is planar; indeed $z=\sqrt3y$.
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