$\textbf{(a)}$ Consider the following graph, the plumb line is not in the same direction as the outward radial from the Earth, so we will define two coordinates system, the primed and the unprimed, as shown. The angle between them is called the angular deviation. The small parameters which govern the approximations that need to be made to find the southerly deflection of a falling particle are:

$$\begin{align} \delta \equiv \frac{h}{R} \qquad \alpha \equiv \frac{R \omega^{2}}{g_{0}} \qquad \delta \alpha=\frac{h \omega^{2} }{ g_{0}}\end{align}$$

where $h$ is the height of fall, $R$ is the radius of Earth, $R \omega^{2}$ is the centrifugal force, and $g_{0}$ purely gravitational force. The angular deviation is given by (from previous problem):

$$\begin{align} \varepsilon \equiv \frac{R \omega^{2}}{g} \sin (\lambda) \cos (\lambda) \qquad g=g_{0}-R \omega^{2} \cos ^{2}(\lambda)\end{align}$$

the unprimed coordinates can be described in terms of the primed coordinates as:

$$\begin{align} x&=x^{\prime} \cos (\varepsilon)+z^{\prime} \sin (\varepsilon) \\ y&=y^{\prime}\\ z&=-x^{\prime} \sin (\varepsilon)+z^{\prime} \cos (\varepsilon) \end{align}$$

The Coriolis force is given by:

$$\mathbf{F}_{X}=-2 m \boldsymbol{\omega} \times \mathbf{v}^{\prime}$$

thus, the acceleration due to this force is:

$$\mathbf{a}_{X}=-2 \boldsymbol{\omega} \times \mathbf{v}^{\prime}$$

but,

$$\begin{align*} \mathbf{v}^{\prime}&=\dot{x}^{\prime}\mathbf{i} +\dot{y}^{\prime} \mathbf{j} +\dot{z}^{\prime} \mathbf{k}\\ \boldsymbol{\omega}&=-\omega \cos (\lambda) \mathbf{i}+ \omega \sin (\lambda)\mathbf{k}\\ \mathbf{r}^{\prime}&=x^{\prime}\mathbf{i}+y^{\prime}\mathbf{j}+(R+z^{\prime})\mathbf{k}\\ m \mathbf{a}_{f}&=-m g_{0} \mathbf{k} \end{align*}$$