Hint 2

Taking ${\bf e}_1$ to be along a long side and ${\bf e}_2$ to be along a short side of the rectangle, the moment of inertia tensor is

$${\bf I}={Ma^2\over 6}\pmatrix{2&-3&0\cr -3&8&0\cr 0&0&10}.$$

For $\hbox{{\boldmath$\omega$ }}=(\omega/\sqrt 5)(2,1,0)$: (a) ${\bf L}=(ma^2\omega/6\sqrt 5)(1,\,2,\,0)$, (b) $T=Ma^2\omega^2/15$,
(c) $\hbox{{\boldmath$\tau$ }}=(Ma^2\omega^2/10)(0,\,0,\,1)$.