The total moment of inertia of the system is
$$I = {I_1} + {I_2} + {I_3}\,......\left( {\text{i}} \right)$$
Here, $${I_1} = \frac{2}{3}m{r^2}$$
$$\eqalign{ & {I_2} = {I_3} = \frac{2}{3}m{r^2} + m{r^2}\,\,\left[ {{\text{From parallel axis theorem}}} \right] \cr & = \frac{5}{3}m{r^2} \cr} $$
From Eq. (i),
$$\eqalign{ & I = \frac{2}{3}m{r^2} + 2 \times \frac{5}{3}m{r^2} = m{r^2}\left( {\frac{2}{3} + \frac{{10}}{3}} \right) \cr & I = 4m{r^2} \cr} $$

Moment of inertia of the system about $$AX$$  is given by

Moment of inertia \[ = {m_A}r_A^2 + {m_B}r_B^2 + {m_C}r_C^2\,\,\left[ {\begin{array}{*{20}{c}} {{r_A} = 0}\\ {{r_B} = l}\\ {{r_C} = l\sin {{30}^ \circ }} \end{array}} \right]\]
Moment of inertia $$ = m{\left( 0 \right)^2} + m{\left( l \right)^2} + m{\left( {l\sin {{30}^ \circ }} \right)^2}$$
$$ = m{l^2} + \frac{{m{l^2}}}{4} = \frac{5}{4}m{l^2}$$
Alternative
Moment of inertia of a system about a line $$OC$$ perpendicular to $$AB$$  in the plane of $$ABC$$  is

$$\eqalign{ & {I_{CO}} = m \times 0 + m \times {\left( {\frac{l}{2}} \right)^2} + m \times {\left( {\frac{l}{2}} \right)^2} \cr & \therefore {I_{CO}} = \frac{{m{l^2}}}{4} + \frac{{m{l^2}}}{4} = \frac{{m{l^2}}}{2} \cr} $$
Now, by applying parallel axes theorem $${I_{AX}} = {I_{CO}} + M{x^2}$$
where, $$x =$$  distance of $$AX$$  from $$CO$$
$$M =$$  total mass of system
$$\eqalign{ & {I_{AX}} = \frac{{m{l^2}}}{2} + 3m \times {\left( {\frac{l}{2}} \right)^2} \cr & {I_{AX}} = \frac{{m{l^2}}}{2} + \frac{{3m{l^2}}}{4} = \frac{5}{4}m{l^2} \cr} $$

$$\eqalign{ & 10\cos {30^ \circ } - f = 2a\,......\left( {\text{i}} \right) \cr & \tau = I\alpha \cr & \Rightarrow fr = \frac{2}{3} \times 2 \times {r^2} \times \alpha \,......\left( {{\text{ii}}} \right)\,\,\left( {{\text{where }}r{\text{ is radius of sphere}}} \right) \cr} $$

From (i) and (ii), we get
$$\eqalign{ & f = 2\sqrt 3 \,{\text{newton,}} \cr & N = 20 + 10\sin {30^ \circ } = 25 \cr & f = \mu N \Rightarrow \mu = \frac{f}{N} = \frac{{2\sqrt 3 }}{{25}} = 0.08 \times \sqrt 3 \cr} $$

$$\eqalign{ & I = 1.2\;kg\;{m^2},{E_r} = 1500\;J, \cr & \alpha = 25\,rad/{\sec ^2},{\omega _1} = 0,t = ? \cr & {\text{As}}\,{E_r} = \frac{1}{2}I{\omega ^2},\omega = \sqrt {\frac{{2{E_r}}}{I}} \cr & = \sqrt {\frac{{2 \times 1500}}{{1.2}}} = 50\,rad/\sec \cr & {\text{From}}\,{\omega _2} = {\omega _1} + \alpha t \cr & 50 = 0 + 25t,t = 2\,{\text{seconds}} \cr} $$

$${X_{cm}} = \int_0^L {\frac{{\left( {dm} \right)x}}{{\int_0^L {\left( {dm} \right)} }} = \frac{{\int_0^L {\left( {\lambda xdx} \right)x} }}{{\int_0^L {\left( {\lambda xdx} \right)} }} = \frac{{2L}}{3}} $$

$${E_r} = \frac{1}{2}I{\omega ^2} \Rightarrow I = \frac{{2{E_r}}}{{{\omega ^2}}} = \frac{{2 \times 360}}{{30 \times 30}} = 0.8\,kg\,{m^2}$$

Here $$a = \frac{2}{{\sqrt 3 }}R$$   Now, $$\frac{M}{{M'}} = \frac{{\frac{4}{3}\pi {R^3}}}{{{a^3}}}$$
$$ = \frac{{\frac{4}{3}\pi {R^3}}}{{{{\left( {\frac{2}{3}R} \right)}^3}}} = \frac{{\sqrt 3 }}{2}\pi .$$

$$M' = \frac{{2M}}{{\sqrt 3 \pi }}$$
Moment of inertia of the cube about the given axis, $$I = \frac{{M'{a^2}}}{6}$$
$$\eqalign{ & = \frac{{\frac{{2M}}{{\sqrt 3 \pi }} \times {{\left( {\frac{2}{{\sqrt 3 }}R} \right)}^2}}}{6} \cr & = \frac{{4M{R^2}}}{{9\sqrt 3 \pi }} \cr} $$

$${I_{nn'}} = $$   $$M.I.$$  due to the point mass at $$B \,\,+$$  $$M.I.$$  due to the point mass at $$D \,\,+$$  $$M.I.$$  due to the point mass at $$C$$
$$\eqalign{ & {I_{nn'}} = 2 \times m{\left( {\frac{\ell }{{\sqrt 2 }}} \right)^2} + m{\left( {\sqrt 2 \ell } \right)^2} \cr & = m{\ell ^2} + 2m{\ell ^2} \cr & = 3m{\ell ^2} \cr} $$

Magnitude of moment of inertia depends on the distribution of mass taken from the axis.
From the axis $$EG,$$  the distribution of masses is at minimum distance while from the axis $$BD$$  the distribution of masses is at maximum distance. Hence, the moment of inertia is minimum along axis through $$EG.$$

The system of two given particles of masses $${m_1}$$ and $${m_2}$$ are shown in figure.

Initially the centre of mass $${r_{CM}} = \frac{{{m_1}{r_1} + {m_2}{r_2}}}{{{m_1} + {m_2}}}\,......\left( {\text{i}} \right)$$
When mass $${m_1}$$ moves towards centre of mass by a distance $$d,$$ then let mass $${m_2}$$ moves a distance $${d'}$$ away from $$CM$$ to keep the $$CM$$ in its initial position.
So, $${r_{CM}} = \frac{{{m_1}\left( {{r_1} - d} \right) + {m_2}\left( {{r_2} + d'} \right)}}{{{m_1} + {m_2}}}\,......\left( {{\text{ii}}} \right)$$
Equating Eqs. (i) and (ii), we get
$$\eqalign{ & \frac{{{m_1}{r_1} + {m_2}{r_2}}}{{{m_1} + {m_2}}} = \frac{{{m_1}\left( {{r_1} - d} \right) + {m_2}\left( {{r_2} + d'} \right)}}{{{m_1} + {m_2}}} \cr & \Rightarrow - {m_1}d + {m_2}d' = 0 \cr & \Rightarrow d' = \frac{{{m_1}}}{{{m_2}}}d. \cr} $$
NOTE
If both the masses are equal i.e., $${m_1} = {m_2},$$   then second mass will move a distance equal to the distance at which first mass is being displaced.