1. Understand the setup

A point charge $Q$ is placed at the centre of a conducting spherical shell.

• Region I: $r<a$ (inside the cavity)
• Region II: $a<r<b$ (inside the conducting material)
• Region III: $r>b$ (outside the shell)

We must determine whether the electric field is zero or non-zero in each region.

2. Region I: $r<a$

This is the empty cavity containing the point charge at the centre.

Using Gauss's law for a spherical Gaussian surface of radius $r<a$:

$\oint \vec E\cdot d\vec A = \frac{Q_{\text{enc}}}{\varepsilon_0}$

Since the point charge $Q$ is enclosed,

$E(4\pi r^2)=\frac{Q}{\varepsilon_0}$

So,

$E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$

Thus, in Region I,

$E_I\neq 0$

3. Region II: $a<r<b$

This region lies inside the conducting material.

In electrostatic equilibrium, the electric field inside a conductor is always zero.

Therefore,

$E_{II}=0$

4. Region III: $r>b$

Outside the shell, consider a Gaussian surface of radius $r>b$.

The central charge $Q$ induces:

• charge $-Q$ on the inner surface,
• charge $+Q$ on the outer surface, if the shell is initially neutral.

Total enclosed charge for $r>b$ is:

$Q+(-Q)+(+Q)=Q$

Hence by Gauss's law,

$E(4\pi r^2)=\frac{Q}{\varepsilon_0}$

So,

$E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$

Thus,

$E_{III}\neq 0$

5. Compare with options

• A: $E_I=0, E_{II}=0, E_{III}\neq 0$ ❌ since $E_I\neq 0$
• B: $E_I\neq 0, E_{II}=0, E_{III}=0$ ❌ since $E_{III}\neq 0$
• C: $E_I\neq 0, E_{II}=0, E_{III}\neq 0$ ✅
• D: $E_I=0, E_{II}=0, E_{III}=0$ ❌

6. Final answer

The correct option is:

$\boxed{\text{C}}$