The tangent planes along the curve $x(t, const)$ are all equal, where $x$ is the tangent surface of $\alpha$

Here is the Ex.2.4.6 in Do Carmo's Differential geometry of curves and surfaces.

Let $\alpha:I\rightarrow \mathbb R^3$ be a regular parametrized curve with everywhere nonzero curvature. Consider the tangent surface of $\alpha$, with the parametrization defined by $$X(t,v)=\alpha(t)+v\alpha'(t), t\in I, v\neq 0.$$

Show that the tangent planes along the curve $x(t, const)$ are all equal.

I think the proof is straightforward. Let $c$ denote that constant. Then the tangent planes along the curve $x(t, const)$ can be written as

$$[X_t\wedge X_v]\cdot(x-x(t,c))=0$$ Since $$X_t(t,c)=\alpha'(t)+c\alpha''(t), X_v(t,c)=\alpha'(t)$$ then $$[\alpha'(t)\wedge\alpha''(t)]\cdot(x-\alpha(t))=0$$

This is all I can do for this problem. But intuitively, I am afraid the tangent planes $[\alpha'(t)\wedge\alpha''(t)](x-\alpha(t))=0$ are not equal to each other, since $\alpha'(t)\wedge\alpha''(t)$ does not necessarily have a constant direction. What's wrong with it?

You want to be looking at the tangent plane along the rulings, i.e., fixing $t$ and varying $v$. Now your calculation works great (if you fiddle with what should be $c$).