1. **Problem statement:** Given vectors $\mathbf{A} = 2\mathbf{i} - 3\mathbf{j}$, $\mathbf{B} = -\mathbf{j} + \mathbf{k}$, and $\mathbf{C} = \mathbf{i} + \mathbf{j} + \mathbf{k}$, find the scalar triple product $\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})$. 2. **Formula:** The scalar triple product is given by $$\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = \det \begin{pmatrix} A_x & A_y & A_z \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{pmatrix}$$ where $\mathbf{A} = (A_x, A_y, A_z)$, $\mathbf{B} = (B_x, B_y, B_z)$, and $\mathbf{C} = (C_x, C_y, C_z)$. 3. **Step 1: Write components:** $\mathbf{A} = (2, -3, 0)$ $\mathbf{B} = (0, -1, 1)$ $\mathbf{C} = (1, 1, 1)$ 4. **Step 2: Compute $\mathbf{B} \times \mathbf{C}$:** $$\mathbf{B} \times \mathbf{C} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = \mathbf{i}((-1)(1) - (1)(1)) - \mathbf{j}(0 \cdot 1 - 1 \cdot 1) + \mathbf{k}(0 \cdot 1 - (-1) \cdot 1)$$ $$= \mathbf{i}(-1 - 1) - \mathbf{j}(0 - 1) + \mathbf{k}(0 + 1) = -2\mathbf{i} + \mathbf{j} + \mathbf{k}$$ 5. **Step 3: Compute $\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})$:** $$= (2, -3, 0) \cdot (-2, 1, 1) = 2 \times (-2) + (-3) \times 1 + 0 \times 1 = -4 - 3 + 0 = -7$$ 6. **Answer:** $$\boxed{-7}$$ This is the scalar triple product $\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})$.