The position vector of centre of mass
$$\eqalign{ & r = \frac{{{m_1}{r_1} + {m_2}{r_2}}}{{{m_1} + {m_2}}} \cr & = \frac{{1\left( {\hat i + 2\hat j + \hat k} \right) + 3\left( { - 3\hat i - 2\hat j + \hat k} \right)}}{{1 + 3}} \cr & = \frac{1}{4}\left( { - 8\hat i - 4\hat j + 4\hat k} \right) \cr & = - 2\hat i - \hat j + \hat k \cr} $$

COM of $${m_1}$$ and $${m_2}$$ masses lies at
$$r = \frac{{{m_1}{r_1} + {m_2}{r_2}}}{{{m_1} + {m_2}}}$$

∴ Moment of inertia of the point masses about the given axis is
$$\eqalign{ & I = \sum {{m_i}r_i^2 \Rightarrow I = {m_1}r_1^2 + {m_2}r_2^2} \cr & = {m_1}{\left( {\frac{{{m_2}l}}{{{m_1} + {m_2}}}} \right)^2} + {m_2}{\left( {\frac{{{m_1}l}}{{{m_1} + {m_2}}}} \right)^2} \cr & = \frac{{{m_1}{m_2}{l^2}}}{{{{\left( {{m_1} + {m_2}} \right)}^2}}} + \left( {{m_2} + {m_1}} \right) = \frac{{{m_1}{m_2}{l^2}}}{{\left( {{m_1} + {m_2}} \right)}} \cr} $$

The whole mass of the ball will be concentrated at the centre of the ball. All the three balls are identical, i.e., the balls have same mass. On each vertex of equilateral $$\Delta PQR,$$  same mass is kept.

Therefore, centre of mass of the triangle is the centre of mass of the system which is point of intersection of the medians of the triangle.

Moment of inertia of a disc about its diameter is $${I_d} = \frac{1}{4}M{R^2}$$

Now, according to perpendicular axis theorem, moment of inertia of disc about a tangent passing through rim and in the plane of disc is $$I = {I_d} + M{R^2} = \frac{1}{4}M{R^2} + M{R^2} = \frac{5}{4}M{R^2}$$

Angular momentum will be conserved
$${I_1}\omega = {I_1}\omega ' + {I_2}\omega ' = \omega ' = \frac{{{I_1}\omega }}{{{I_1} + {I_2}}}$$

Given, $$r = 5\,cm = 5 \times {10^{ - 2}}m$$
$$T = 0.2\,\pi s$$
We know that acceleration is given by
$$a = r{\omega ^2} = \frac{{4{\pi ^2}}}{{{T^2}}}r = \frac{{4 \times {\pi ^2} \times 5 \times {{10}^{ - 2}}}}{{{{\left( {0.2\,\pi } \right)}^2}}} = 5\,m/{s^2}$$

$$\eqalign{ & \overline x = \frac{{m\left( { - 2b} \right) + 2m\left( { - b} \right) + m \times 0 + 2m\left( b \right)}}{{m + 2m + m + 2m}} = \frac{{ - b}}{3} \cr & {\text{and}}\,\overline y = + b. \cr} $$

For a thin uniform square sheet
$${I_1} = {I_2} = {I_3} = \frac{{m{a^2}}}{{12}}$$

Let $$\sigma $$  be the mass per unit area.

The total mass of the disc $$ = \sigma \times \pi {R^2} = 9M$$
The mass of the circular disc cut
$$ = \sigma \times \pi {\left( {\frac{R}{3}} \right)^2} = \sigma \times \frac{{\pi {R^2}}}{9} = M$$
Let us consider the above system as a complete disc of mass $$9\,M$$  and a negative mass $$M$$  super imposed on it.
Moment of inertia $$\left( {{I_1}} \right)$$   of the complete disc $$ = \frac{1}{2}9M{R^2}$$   about an axis passing through $$O$$  and perpendicular to the plane of the disc.
$$M.I.$$  of the cut out portion about an axis passing through $$O'$$  and perpendicular to the plane of disc $$ = \frac{1}{2} \times M \times {\left( {\frac{R}{3}} \right)^2}$$
$$\therefore \,\,M.I.\left( {{I_2}} \right)$$   of the cut out portion about an axis passing through $$O$$ and perpendicular to the plane of disc
$$ = \left[ {\frac{1}{2} \times M \times {{\left( {\frac{R}{3}} \right)}^2} + M \times {{\left( {\frac{{2R}}{3}} \right)}^2}} \right]$$         [Using perpendicular axis theorem]
$$\therefore $$  The total $$M.I.$$  of the system about an axis passing through $$O$$  and perpendicular to the plane of the disc is $$I = {I_1} + {I_2}$$
$$\eqalign{ & = \frac{1}{2}9M{R^2} - \left[ {\frac{1}{2} \times M \times {{\left( {\frac{R}{3}} \right)}^2} + M \times {{\left( {\frac{{2R}}{3}} \right)}^2}} \right] \cr & = \frac{{9M{R^2}}}{2} - \frac{{9M{R^2}}}{{18}} \cr & = \frac{{\left( {9 - 1} \right)M{R^2}}}{2} \cr & = 4\,M{R^2} \cr} $$

$${I_{nn'}} = \frac{1}{{12}}M\left( {{a^2} + {a^2}} \right) = \frac{{M{a^2}}}{6}$$

Also, $$DO = \frac{{DB}}{2} = \frac{{\sqrt 2 a}}{2} = \frac{a}{{\sqrt 2 }}$$
According to parallel axis theorem
$$\eqalign{ & {I_{mm'}} = {I_{nn'}} + M{\left( {\frac{a}{{\sqrt 2 }}} \right)^2} = \frac{{M{a^2}}}{6} + \frac{{M{a^2}}}{2} \cr & = \frac{{M{a^2} + 3M{a^2}}}{6} \cr & = \frac{2}{3}M{a^2} \cr} $$