Let \( P, Q \), and \( R \) be points on a plane given by \( P=(-1,-4,-3) \), \( Q=(2,4,3) \), and \( R=(-2,1,0) \). Find the cross product \( P Q \times P R \). \[ P Q \times P R=[\text { Ex: } 42, \square, 23] \]

To find the cross product \( P Q \times P R \), we first need to determine the vectors \( P Q \) and \( P R \). The vector \( P Q \) is given by: \[ P Q = Q - P = (2 - (-1), 4 - (-4), 3 - (-3)) = (2 + 1, 4 + 4, 3 + 3) = (3, 8, 6). \] The vector \( P R \) is given by: \[ P R = R - P = (-2 - (-1), 1 - (-4), 0 - (-3)) = (-2 + 1, 1 + 4, 0 + 3) = (-1, 5, 3). \] Next, we compute the cross product \( P Q \times P R \): \[ P Q = (3, 8, 6), \quad P R = (-1, 5, 3). \] Using the formula for the cross product: \[ P Q \times P R = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 8 & 6 \\ -1 & 5 & 3 \end{vmatrix}. \] Calculating the determinant: \[ = \mathbf{i} \begin{vmatrix} 8 & 6 \\ 5 & 3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & 6 \\ -1 & 3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & 8 \\ -1 & 5 \end{vmatrix}. \] Computing the 2x2 determinants: 1. \( \begin{vmatrix} 8 & 6 \\ 5 & 3 \end{vmatrix} = (8)(3) - (6)(5) = 24 - 30 = -6 \). 2. \( \begin{vmatrix} 3 & 6 \\ -1 & 3 \end{vmatrix} = (3)(3) - (6)(-1) = 9 + 6 = 15 \). 3. \( \begin{vmatrix} 3 & 8 \\ -1 & 5 \end{vmatrix} = (3)(5) - (8)(-1) = 15 + 8 = 23 \). Putting it all together: \[ P Q \times P R = -6\mathbf{i} - 15\mathbf{j} + 23\mathbf{k} = (-6, -15, 23). \] Thus, the cross product \( P Q \times P R \) is: \[ \boxed{[-6, -15, 23]}. \]