$\alpha(t)=(\cos(t),\sin(t),f(t))$ must be a flat curve. Find the general form of $f(t)$, and find the special form of the function so that it lays on the surface $Ax+By+Cz+D=0.$
Well for the curve to be flat, it's torsion must be $0$. So I fount that:
$$f'''(t)+f'(t)=0$$ and the solution of this differential equation is (if I'm not mistaken): $$f(t)=c_1+c_2e^{it}+c_3te^{it}$$
How do I know which plane it lays on? Can I find this plane by just having the vector normal vector of the plane, (which is, again, if im not mistaken)$$\alpha'(0)\times\alpha''(0)$$ and a certain point, say $\alpha(0)$. But again how do I put this all together? (That is if I am correct so far in my calculations, and thoughts)
Since it is to be flat it must lie in a plane and a plane is given by an equation on the form $Ax+By+Cz+D=0$. This means that we must have:
$$A\cos t+ B\sin t+Cf(t) + D = 0$$
Which means that:
$$f(t) = -{A\cos t+B\sin t+D\over C}$$
The only complication is for planes where $C=0$, but then we would require that $A\cos t+B\sin t+D = 0$ which would require $A=B=0$ which would degenerate the equation to not be a plane.
If you want (to reduce the number of parameters and simplify the expression) you can rewrite the formula for $f$ by setting $\alpha = -A/C$, $\beta = -B/C$ and $\delta = -D/C$ and get:
$$f(t) = \alpha \cos t + \beta \sin t + \delta$$