So I was solving this larger question and after solving everything, I obtained these data. All derivatives are wrt time unless specified. $f(P)$ represents the position vector of the point $$f(P)=x\hat{i}+y\hat{j}+z\hat{k}$$ $$f'(P)=\vec{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$$ $$\vec{f''(P)}=\vec{A}=\vec{v}\times\vec{B}=\vec{v}\times({a\hat{i}+b\hat{j}+c\hat{k}})$$
I proceed as: Then $f''(P)$ is given by $\left|\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ v_x& v_y & v_z \\ a & b & c \\ \end{matrix}\right|$ Therefore $$\vec{A}=(cv_y-bv_z)\hat{i}-(cv_x-av_z)\hat{j}+(bv_x-av_y)\hat{k}$$ where $a,b,c$ are constants. $$av_x+bv_y+cv_z=\alpha\neq 0$$
$$\cos\theta=\left(\cfrac{\alpha}{\sqrt{a^2+b^2+c^2}×\sqrt{v_x^2+v_y^2+v_z^2}}\right)$$
$$\frac{dv_x}{dt}=cv_y-bv_z$$ $$\frac{dv_y}{dt}=av_z-cv_x$$ $$\frac{dv_z}{dt}=bv_x-av_y$$
Component of $ f'(P) $ along $$\vec{B}=\left(\cfrac{\alpha}{a^2+b^2+c^2}(a\hat{i}+b\hat{j}+c\hat{k})\right)$$
How do I proceed further to show this is a circular helix? In other words when $\vec{v}\times \vec{B}=\vec{A}$ and $\vec{v}.\vec{B}\neq 0$ how to prove that the path is helical? Why is path helical when acceleration is cross product of velocity and a given fixed vector?
In case anyone wants to know the context, This is a proof to show that a charged particle moving at an angle$\neq \pi/2$ to magnetic field experiences a force that moves it in a helical path.