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We need to use the following properties of determinants:\n\n1. $\det(kA) = k^n \det(A)$, where $A$ is an $n \times n$ matrix.\n2. $\det(AB) = \det(A) \det(B)$.
We are given $\det(A) = 3$ and $\det(B) = -4$. We want to find $\det(-6AB)$.\n\nUsing the properties, we have:\n$$\det(-6AB) = \det(-6I \cdot AB) = \det(-6I) \det(A) \det(B)$$\nwhere $I$ is the identity matrix of order 3.\n\nSince $-6I$ is a $3 \times 3$ matrix, $\det(-6I) = (-6)^3 = -216$.\n\nTherefore,\n$$\det(-6AB) = (-6)^3 \det(A) \det(B) = (-216) \cdot (3) \cdot (-4)$$\n$$\det(-6AB) = -216 \cdot (-12) = 2592$$
Final Answer: 2592