1. The problem defines a custom operation $X$ in equation I: $$a X b = \begin{cases} 2a + b, & \text{if } a \leq b \\ a^2 - b^2, & \text{if } a > b \end{cases}$$ 2. We need to find the value of $(-2) X (4 X 3)$ in equation II. 3. First, calculate the inner operation $4 X 3$: Since $4 > 3$, use the second case: $$4 X 3 = 4^2 - 3^2 = 16 - 9 = 7$$ 4. Now calculate $(-2) X 7$: Since $-2 \leq 7$, use the first case: $$(-2) X 7 = 2(-2) + 7 = -4 + 7 = 3$$ 5. Therefore, the value of $(-2) X (4 X 3)$ is $3$. 6. The correct answer is A) 3.