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This website here: https://www.chegg.com/learn/calculus/calculus/moment-of-inertia-about-the-origin
Shows the following: 
How is it possible to define a mass moment of inertia about the origin which is a point and not an axis. What is the direction of rotation referred to?
What situation would this represent?
The original problem seems to ask about "the axis at the origin perpendicular to the plane", but still, we do have a definition of inertia about a point instead of an axis
consider total angular momentum of a single particle as a vector, denoted by $\vec{L}$ , by definition $$ \vec{L} \equiv \vec{r} \times \vec{P} $$ componentwise: $$ \begin{aligned} L^i & =(r \times P)^i \\ & =\varepsilon^{i j k} r_j P_k \end{aligned} $$ but $ \vec{P}=(\vec{\omega} \times \vec{r})m $ , so its $k$th component is: $$ P_k=(\varepsilon_{klm}\omega^l r^m)m $$ so we have: $$ \begin{aligned} L^i & =\varepsilon^{i j k} \varepsilon_{k l m} r_j r^m \omega^l m \\ & =\left(\delta_l^i \delta_m^j-\delta_m^i \delta_l^j\right) r_j r^m \omega^l m \\ & =\left(r_j r^j \delta_l^i-r^i r_l\right) \omega^l m \\ & =\left(|r|^2 \delta_l^i-r^i r_l\right) m \omega^l \end{aligned} $$ so it's natural to introduce a tensor(called inertia tensor) by: $$ I_j^i=\left(|r|^2 \delta_j^i-r^i r_j\right) m $$ then we can express angular momentum in a tensorial form: $$ L^i=I_j^i \omega^j $$
for continuous distribution, simply replace m by $\rho d^3x$ and do integral
