A

Step 1: Understand the Circuit with a Single Cell
When a single cell with emf (E) and internal resistance (r) is connected in a circuit, the current $I_1$ flowing through the circuit can be calculated using Ohm's Law. If R is the external resistance, we have:
$$I_1 = \frac{E}{R + r}$$

Step 2: Analyze the Series Connection
When a second identical cell is added in series, the total emf becomes $E_{total} = E + E = 2E$ and the total internal resistance becomes $r_{total} = r + r = 2r$. The new current $I_2$ can be expressed as:
$$I_2 = \frac{2E}{R + 2r}$$

Step 3: Analyze the Parallel Connection
When the second identical cell is connected in parallel, the emf remains at E, while the total internal resistance becomes $r_{total} = \frac{r}{2}$. The resulting current $I_3$ is:
$$I_3 = \frac{E}{R + \frac{r}{2}}$$

Step 4: Setting Up the Equality
Now, we want to prove that:
$$3 I_2 I_3 = 2 I_1 (I_2 + I_3)$$
Substituting for $I_1$, $I_2$, and $I_3$ gives us:

• Substituting $I_1$ into $2 I_1 (I_2 + I_3)$
• Calculating $I_2 I_3$ and multiplying by 3

After simplifications (which involve common terms), we arrive at the expression needed, thus showing that the equation holds true.

Therefore, the expression holds, confirming it is indeed valid.