Consider a body whose centre of mass is situated at the origin,...

The moments of inertia of a body about the $X$, $Y$, and $Z$ axes are given by:

$I_x = \sum m_i (y_i^2 + z_i^2)$

$I_y = \sum m_i (x_i^2 + z_i^2)$

$I_z = \sum m_i (x_i^2 + y_i^2)$

Adding the moments of inertia about the $X$ and $Y$ axes, we get:

$I_x + I_y = \sum m_i (x_i^2 + y_i^2 + 2z_i^2)$

$I_x + I_y = \sum m_i (x_i^2 + y_i^2) + 2 \sum m_i z_i^2$

$I_x + I_y = I_z + 2 \sum m_i z_i^2$

The perpendicular axis theorem, which states $I_z = I_x + I_y$, is only applicable for a two-dimensional planar body (lamina) lying entirely in the $XY$ plane, where $z_i = 0$ for all particles.

The problem specifies that one of the particles is located at $(0, 0, a)$, which implies $z_i \neq 0$. Thus, the body is three-dimensional and the perpendicular axis theorem cannot be applied.

Since the complete mass distribution of the body is not provided, the value of $\sum m_i z_i^2$ is unknown. Therefore, the moment of inertia about the $Z$-axis, $I_z$, cannot be determined from the given information.