C

Step 1: Understanding Capacitance
When a dielectric material is inserted between the plates of a capacitor, the capacitance increases. The capacitance $C$ of a parallel plate capacitor with a dielectric is given by:
$C = \frac{\varepsilon A}{d}$, where $\varepsilon = \varepsilon_0 \cdot K$, $K$ is the dielectric constant, $A$ is the area of the plates, and $d$ is the distance between them.
Step 2: Effect on Potential
As the capacitance increases, for a given charge $Q$ on the plates, the potential $V$ across the capacitor is given by:
$V = \frac{Q}{C}$.
Since capacitance $C$ increases, the potential $V$ across the plates will decrease (for constant charge). However, this only applies if the capacitor is isolated. If the capacitor is connected to a battery, the potential will remain constant as the charge adjusts.
Step 3: Effect on Potential Energy
The potential energy $U$ stored in a capacitor is given by:
$U = \frac{1}{2} CV^2$ or $U = \frac{Q^2}{2C}$.
If capacitance increases (even considering constant charge), potential energy increases because the energy stored in the electric field increases due to the presence of the dielectric.
Conclusion: When a dielectric is placed between the plates, the capacity increases, the potential decreases (if isolated), and the potential energy stored increases. Thus, the answer considering the general case where we primarily look at the effect of the dielectric on the capacity and potential energy leads us to Option C: Increase, increase, increase.