By Frenet-Serret formula, we have $B'(s)=\tau(s) N(s)$, where $B(s)$ is binormal and $N(s)$ is normal. Since $\tau(s)\equiv 0$, we have $B'(s)=0$ which implies that $B(s)$ must be a constant vector.
Now choose any point $\alpha(s_0)$ on the curve $\alpha$, consider the function given by $f(s)=(\alpha(s)-\alpha(s_0))\cdot B(s)$. Then $f'(s)=\alpha'(s)\cdot B(s)+(\alpha(s)-\alpha(s_0))\cdot B'(s)=T(s)\cdot B(s)+(\alpha(s)-\alpha(s_0))\cdot 0=0$ where $T(s)$ is the tangent and is orthogonal to $B(s)$. Hence, $f(s)$ must be a constant function, i.e. $(\alpha(s)-\alpha(s_0))\cdot B(s)= c\mbox{ for all }s$ for some constant $c$. On the other hand, when $s=s_0$, the left hand side is zero. Therefore, $c=0$, i.e. $(\alpha(s)-\alpha(s_0))\cdot B(s)= 0\mbox{ for all }s.$ Recall that $B(s)$ is a constant vector, i.e. $B(s)=B$ for some constant vector $B$. Hence, we have $(\alpha(s)-\alpha(s_0))\cdot B= 0\mbox{ for all }s,$ which implies that $\alpha(s)$ lies on the plane with $B$ as the unit normal to the plane.