Suppose you have three points of the form $\left( \rho \cos\theta_i\cos\phi, \rho\sin\theta_i\cos\phi, \rho \sin\phi\right).$ In other words, these three points are equidistant from the origin and are parallel to the $xy-$plane.
Given a unit vector $\vec{n} \neq \vec{k} = \left<0, 0, 1\right>$, I'd like to find the functions $\Delta\theta_i = \Delta\theta_i(\theta_i, \phi, \vec{n})$ and $\Delta\phi_i = \Delta\phi_i(\theta_i, \phi, \vec{n})$ such that the points
$$\left( \rho \cos(\theta_i + \Delta\theta_i)\cos(\phi + \Delta\phi_i), \rho\sin(\theta_i+\Delta\theta_i)\cos(\phi + \Delta\phi_i), \rho \sin(\phi + \Delta\phi_i)\right)$$
exist in the plane normal to $\vec{n}$ and maintain the same relative distances (i.e., the transformation is a rotation).
What is a good way to determine these functions?
First note that you have an unorthodox convention for your spherical coordinate system ... $\cos \phi \leftrightarrow \sin \phi$ is usual.
To construct the required rotation matrix it is (probably) easiest to break it into two rotation matricies; First a rotation about the $y$-axix & then a rotation about the $z$-axis. We require the vector $(0,0,1)$ to be mapped to $(v_1,v_2,v_3)$.
We have \begin{eqnarray*} v_1&=&\cos(\theta') \cos(\phi') \\ v_2&=&\sin(\theta') \cos(\phi') \\ v_3&=& \sin(\phi') \\ \end{eqnarray*} Inverting these gives \begin{eqnarray*} \theta' = \tan^{-1}\left(\frac{v_2}{v_1}\right) \\ \phi'=\tan^{-1}\left(\frac{v_3}{\sqrt{v_1^2 +v_2^2}}\right) \\ \end{eqnarray*} A rotation about the $y$-axis through an angle $\phi'$ is given by \begin{eqnarray*} \begin{bmatrix} \sin(\phi') && 0 && \cos(\phi') \\ 0 && 1 && 0 \\ -\cos(\phi') && 0 && \sin(\phi') \end{bmatrix} \end{eqnarray*} And a rotation about the $z$-axis through an angle $\theta'$ is given by \begin{eqnarray*} \begin{bmatrix} \cos(\theta') && -\sin(\theta') && 0 \\ \sin(\theta') && \cos(\theta') && 0 \\ 0 && 0 && 1 \end{bmatrix} \end{eqnarray*}
We just need to compose these to get \begin{eqnarray*} \begin{bmatrix} \cos(\theta') && -\sin(\theta') && 0 \\ \sin(\theta') && \cos(\theta') && 0 \\ 0 && 0 && 1 \end{bmatrix} \begin{bmatrix} \sin(\phi') && 0 && \cos(\phi') \\ 0 && 1 && 0 \\ -\cos(\phi') && 0 && \sin(\phi') \end{bmatrix} = \color{red}{\begin{bmatrix} ? && ? && ? \\ ? && ? && ? \\ ? && ? && ? \end{bmatrix}}. \end{eqnarray*}