disclaimer: my math is sketchy at best AND english is not my first language, so... i might have some issues naming things - but i'll try my best to be clear :)
given this parametric curve:
anyway, its equation is (sorry if wrong notation):
$$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3\cos(t) + \cos(t)\cos(6t) \\ 3\sin(t) + \sin(t)\cos(6t) \\ \sin(6t) \end{bmatrix} ,t=0...2\pi $$
now, if a circle goes along that curve, and it stays perpendicular to the curve as it goes along, it makes... what? how would you call that thing? maybe a "twisted torus"?
and the question is: what is the parametric equation of the surface of that "twisted torus"?
i hope this question makes sense – it does, in my mind. anyway, thanks for your time!
Let's call your things $x(t), y(t), z(t)$, OK? Let $$ v(t) = \begin{bmatrix} x'(t) \\ y'(t) \\ z'(t) \end{bmatrix} \\ T(t) = v(t) / \| v(t) \|. $$ Then $T$ will be tangent to your curve at each time $t$. Do the same thing with $x'', y'', z''$ to get $$ w(t) = \begin{bmatrix} x''(t) \\ y''(t) \\ z''(t) \end{bmatrix} \\ u(t) = w(t) - ( u(t) \cdot T(t) ) T(t) \\ N(T) = u(t) / \| u(t) \|. $$ Then $N(t)$ will be perpendicular to $T(t)$ at each point. Finally, let $$ B(t) = T(t) \times N(t). $$
Now: $$ S(s, t) = T(t) + r \sin(s) N(t) + r \cos(s) B(t) $$ will, as $s$ ranges from $0$ to $2\pi$, and $t$ ranges over its usual range, and for small enough values of $r$ (like $r = 0.1$) sweep out a tube around your curve.
What I've done is construct for you the Frenet-Serret frame for the curve, which assumes that the curvature (the length of the vector $u(t)$) is never zero; if it is, you have to use the "Bishop frame", which ... takes more work to write out. I think that for your curve, the F-S frame will work fine.