As we know that, the kinetic energy of a rotating body,
$$KE = \frac{1}{2}I{\omega ^2} = \frac{1}{2}\frac{{{I^2}{\omega ^2}}}{I} = \frac{{{L^2}}}{{2I}}$$
Also, angular momentum, $$L = I\omega $$
Thus, $${K_A} = {K_B}$$
$$\eqalign{ & \Rightarrow \frac{1}{2}\frac{{L_A^2}}{{{I_A}}} = \frac{1}{2}\frac{{L_B^2}}{{{I_B}}} \cr & \Rightarrow {\left( {\frac{{{L_A}}}{{{L_B}}}} \right)^2} = \frac{{{I_A}}}{{{I_B}}} \Rightarrow \frac{{{L_A}}}{{{L_B}}} = \sqrt {\frac{{{I_A}}}{{{I_B}}}} \cr & L \propto \sqrt I \cr & \therefore {L_A} < {L_B}\,\,\left[ {\because {I_B} > {I_A}} \right] \cr} $$

Moment of Inertia depend upon mass and distribution of masses as $$I = \Sigma m{r^2}.$$
Further, as the distance of masses is more, more is the moment of Inertia.

If we choose $$BC$$ as axis. Distance is maximum.
Hence, Moment of Inertia is maximum.
$$\therefore {I_2} > {I_1},{I_2} > {I_3}$$

Angular momentum,
$$\eqalign{ & {L_0} = mvr\sin {90^ \circ } \cr & = 2 \times 0.6 \times 12 \times 1 \times 1\,\,\left[ {{\text{As}}\,V = r\omega ,\sin {{90}^ \circ } = 1} \right] \cr & {\text{So,}}\,\,{L_0} = 14.4\,kg{m^2}/s \cr} $$

$$\tau = r \times F,$$   where $$r =$$  position vector
$$F = {\text{force}} \Rightarrow \tau = \left| r \right| \cdot \left| F \right|\sin \theta $$
Torque is perpendicular to both $$r$$ and $$F.$$ So, dot product of two vectors will be zero.
$$\therefore \tau \cdot r = 0 \Rightarrow F \cdot \tau = 0$$