= Solution
The <parallel axis theorem> relates the <moment of inertia> about an axis through the <centre of mass> to that about a parallel axis displaced by perpendicular distance $d$:
$$
\boxed{I=I_G+Md^2.}
$$
The reference axis must pass through the <centre of mass>. To see why, write the displacement from it as $\mathbf r=\mathbf R+\mathbf d$ in the plane perpendicular to the axes. Expanding $\int|\mathbf R+\mathbf d|^2\,dm$ gives $I_G+Md^2$ because $\int\mathbf R\,dm=0$.
For a thin planar <mass distribution>, let $x,y$ be perpendicular axes in its plane and $z$ the normal axis through \b[their same intersection]. The <perpendicular axis theorem> states
$$
\boxed{I_z=I_x+I_y.}
$$
Indeed, $I_z=\int(x^2+y^2)\,dm$, while $I_x=\int y^2\,dm$ and $I_y=\int x^2\,dm$. Their common intersection need not be the <centre of mass>, but an arbitrary three-dimensional body does not satisfy the planar equality.
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