1. First, evaluate $\det(2A^T)$ given $\det(A) = 3$.\n Since $A$ is a $3 \times 3$ matrix, scaling by 2 multiplies the determinant by $2^3 = 8$. Also, the determinant of the transpose $A^T$ equals $\det(A)$.\n Therefore, \n$$\det(2A^T) = 2^3 \times \det(A^T) = 8 \times 3 = 24.$$\n 2. Next, evaluate $\det(EA)$ where $E$ is an elementary matrix of type I, and $\det(A) = -120$.\n Type I elementary matrices correspond to swapping two rows, which multiplies the determinant by $-1$.\n Thus, \n$$\det(E) = -1.$$\n Therefore, \n$$\det(EA) = \det(E) \cdot \det(A) = -1 \times (-120) = 120.$$