Let the common mass of each body be $M$ and common radius be $R$.

We need the total moment of inertia of the system about the axis $PQ$.

The figure is not shown here, but from the standard arrangement for this question, the three bodies are touching each other in a line and the axis $PQ$ passes through the center of the middle body, perpendicular to the line joining centers. Hence the centers are at perpendicular distances:

• disc $A$: $2R$ from $PQ$
• solid sphere $B$: $0$
• spherical shell $C$: $2R$ from $PQ$

We are also given that $I$ is the moment of inertia of the disc about its diameter.

1. Reference quantity $I$

For a disc about its diameter,

2. Moment of inertia of each body about its own central axis parallel to $PQ$

Since $PQ$ is parallel to a diameter of the disc and to corresponding central axes of the spheres:

1. Disc $A$ about a diameter:

2. Solid sphere $B$ about any diameter:

3. Spherical shell $C$ about any diameter:

3. Apply parallel axis theorem

For disc $A$ and shell $C$, distance from their centers to $PQ$ is $2R$.

Disc $A$

Solid sphere $B$

Its center lies on $PQ$, so no shift:

Spherical shell $C$

4. Total moment of inertia

Taking LCM $=60$,

So,

5. Express in terms of $I$

Since

we get

Comparing with

we get

6. Comparison with stored answer

Stored correct answer is $199$, but the derived answer is $559$.

So I do not agree with the stored answer. The stored answer may correspond to a different axis placement than the usual one inferred from the standard figure, but for the standard touching arrangement with $PQ$ through the middle body, the correct value is $559$.