To solve this system of equations using matrices and determinants, we can write the system in matrix form:

\begin{bmatrix}2 & -3 & 4 \\ -3 & 4 & 2 \\ 4 & -2 & -3\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}-9 \\ -12 \\ -3\end{bmatrix}

We can then use the formula for the inverse of a matrix to solve for the variables:

$$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = A^{-1} \begin{bmatrix}-9 \\ -12 \\ -3\end{bmatrix}$$

where $A^{-1}$ is the inverse of the coefficient matrix:

$$A^{-1} = \frac{1}{\det(A)} \begin{bmatrix}c_{11} & c_{21} & c_{31} \\ c_{12} & c_{22} & c_{32} \\ c_{13} & c_{23} & c_{33}\end{bmatrix}$$

where $\det(A)$ is the determinant of the coefficient matrix, and $c_{ij}$ is the $(i,j)$th cofactor of the matrix. Using the formula for the determinant of a 3x3 matrix, we have:

$$\det(A) = 2(4(-3) - 2(-2)) - (-3)(-3(-3) - 2(4)) + 4(-3)(-3) = -119$$

Using the formula for the cofactors, we have:

$$c_{11} = 4, \quad c_{12} = -2, \quad c_{13} = -3$$
$$c_{21} = -(-3), \quad c_{22} = 2(2), \quad c_{23} = -4$$
$$c_{31} = -2(2), \quad c_{32} = -(4), \quad c_{33} = 2(-3)$$

Therefore, we have:

$$A^{-1} = \frac{1}{-119} \begin{bmatrix}4 & 6 & -2 \\ 3 & 8 & 4 \\ 8 & -6 & 6\end{bmatrix}$$

Multiplying $A^{-1}$ by the right-hand side of the equation, we obtain:

$$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{-119} \begin{bmatrix}4 & 6 & -2 \\ 3 & 8 & 4 \\ 8 & -6 & 6\end{bmatrix} \begin{bmatrix}-9 \\ -12 \\ -3\end{bmatrix} = \begin{bmatrix}1 \\ -1 \\ 1\end{bmatrix}$$

Therefore, the solution to the system of equations is $x=1$, $y=-1$, and $z=1$.