I've solved this problem, but why is this differentiable? Let $\alpha:\mathbb R\rightarrow \mathbb R^3$ be a smooth curve (i.e., $\alpha \in C^\infty(\mathbb R)$). Suppose there exists $X_0$ such that for every normal line to $\alpha$, $X_0$ belongs to it. Show that $\alpha$ is part of a circumference.
Part of solution: Let $n(s)$ be the normal vector to $\alpha$ on $s$. For every $s \in \mathbb R$ there exists an unique real number $\lambda$ such that $\alpha(s)+\lambda n(s)=X_0$.
Now let $\lambda(s)$ be the function that associates every real number $s$ to this number.

I've finished this question, but I have derivated $\lambda(s)$ and I don't know why I can do it. Can someone tell me why is $\lambda$ differentiable?

You can write
$$\lambda(s)=\langle X_0-\alpha(s), n(s)\rangle$$
supposing $\|n(s)\|=1$ (otherwise you divide by this norm).
That expression is obtained with sums and products of smooth functions (at most divisions by non-vanishing smooth functions), hence it is smooth.
