First, some note on this paper: from my cursory reading, it always assumes zero atmsohperic wind relative to the ground. That's why the wind axis represents the true motion of the aircraft. In general, this is not true, therefore $\psi_W$ and $\theta_W$ are not the track and flight path angles.
Now onto your question. When you consider zero wind, then the angular difference between the y wind axis ($\hat{j}_{W}$) and the y body axis ($\hat{j}_{B}$) is the sideslip angle, $\beta$. This is consistent when you consider flow incidences as 3-2-1 Euler rotation from the body axis, where yaw (sideslip) is the last rotation:
$$C_{BW} = \begin{bmatrix}\cos\alpha\cos\beta & -\cos\alpha\sin\beta & -\sin\alpha \\
\sin\beta & \cos\beta & 0 \\
\sin\alpha\cos\beta & -\sin\alpha\sin\beta & \cos\alpha
\end{bmatrix}$$
Mathematically, this can be immediately seen when you rewrite the rotation matrix in its equivalent direction cosine form:
$$C_{BW} = \begin{bmatrix}\hat{i}_W \cdot \hat{i}_B & \hat{j}_W \cdot \hat{i}_B & \hat{k}_W \cdot \hat{i}_B \\
\hat{i}_W \cdot \hat{j}_B & \hat{j}_W \cdot \hat{j}_B & \hat{k}_W \cdot \hat{j}_B \\
\hat{i}_W \cdot \hat{k}_B & \hat{j}_W \cdot \hat{k}_B & \hat{k}_W \cdot \hat{k}_B\end{bmatrix}$$
So: $\cos{\beta}=\hat{j}_W \cdot \hat{j}_B$.
The gradient of the Cartesian equation of a plane is its unit normal, when normalized. For the $x_W z_W$ and $x_B z_B$ planes, the normal is simply their y axis (i.e. $\hat{j}_W$ and $\hat{j}_B$, respectively.
Lastly, eqn (72) can be worked out when $\begin{bmatrix}x & y & z\end{bmatrix}^T$ in the planar equations are in the inertial frame ($_V$), and transformed to the respective frames. The coefficients can then be derived and massaged to obtain (72).
