Step 1 (Combine two first): Instead of adding all three at once, first replace \(C_1\) and \(C_2\) in parallel by a single capacitor. Both share the same voltage \(V\), so their charges are \(Q_1 = C_1 V\) and \(Q_2 = C_2 V\), giving a combined charge \(Q_1 + Q_2 = (C_1 + C_2)V\). Hence their equivalent value is
\[ C_{12} = \frac{Q_1 + Q_2}{V} = C_1 + C_2. \]
Step 2 (Bring in the third): Now \(C_{12}\) and \(C_3\) are themselves in parallel across the same \(V\). Repeating the same reasoning,
\[ C_p = C_{12} + C_3 = (C_1 + C_2) + C_3. \]
Step 3 (Result): Therefore
\[ C_p = C_1 + C_2 + C_3. \]
Step 4 (Physical check): Capacitance measures charge stored per unit voltage. Putting plates side by side in parallel effectively enlarges the total plate area facing each other, and since \(C \propto\) area, the capacitances simply add. This also explains why the parallel result exceeds every individual capacitor.
\[\boxed{C_p = C_1 + C_2 + C_3}\]