From looking at your diagram, it appears to me that the transformation takes the unit vector $(i_x,i_y,i_z)$ and rotates it to point to $(f_x,f_y,f_z)$:
$$\begin{pmatrix}c_\Psi&s_\Psi&0\\\\-s_\Psi&c_\Psi&0\\\0&0&1\end{pmatrix} \begin{pmatrix}1&0&0\\\0&c_\theta&-s_\theta\\\0&s_\theta&c_\theta\end{pmatrix} \begin{pmatrix}c_\phi&-s_\phi&0\\\s_\phi&c_\phi&0\\\0&0&1\end{pmatrix} \begin{pmatrix}i_x\\\i_y\\\i_z\end{pmatrix} = \begin{pmatrix}f_x\\\f_y\\\f_z\end{pmatrix} $$
where $c_\phi = \cos(\phi), s_\phi = \sin(\phi)$, etc.
It isn't hard to solve for $\phi,\theta$ and $\Psi$ if you assume that your camera begins by pointing in a convenient direction such as $(i_x,i_y,i_z) = (0,0,1)$.