A point charge Q is placed inside a cavity within a solid isolated...

Point A is located inside the cavity where a point charge $Q$ is present. The electric field inside the cavity is the superposition of the field due to the charge $Q$ and the induced charge on the inner surface of the cavity. This resultant field is non-zero, so $E_A \neq 0$.

The point charge $Q$ inside the cavity induces a charge $-Q$ on the inner surface of the cavity and a charge $+Q$ on the outer surface of the isolated conducting sphere. Because the outer surface of the sphere is perfectly spherical and there are no external electric fields, the induced charge $+Q$ distributes itself uniformly over the outer surface.

For points outside the conducting sphere, the electric field is determined solely by this uniformly distributed charge on the outer surface. By Gauss's law, the field outside is identical to that of a point charge $+Q$ located at the center of the sphere.

Since points B and C are outside the sphere and are at the same distance from the center of the solid sphere, the magnitudes of the electric field at these points must be equal. Therefore, $E_B = E_C$.

Combining these observations, we get $E_A \neq 0$ and $E_B = E_C$.

Answer: $E_A\neq0,\,E_B=E_C$