Find the equation of the plane through the points \( P=(-2,2,4) \), \( Q=(-1,0,-3) \), and \( R=(1,3,-4) \) given \( P Q=(1,-2,-7) \) and \( P R=(3,1,-8) \). Use the point \( P \) for \( \left(x_{0}, y_{0}, z_{0}\right) \). Ex.2x+3y-4z\( +44=0 \) Enter as: \( A x+B y+C z \)

To find the equation of the plane through the points \( P, Q, R \), we can use the normal vector, which is obtained by taking the cross product of vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \). 1. **Calculate the cross product**: \[ \overrightarrow{PQ} = Q - P = (-1 + 2, 0 - 2, -3 - 4) = (1, -2, -7) \] \[ \overrightarrow{PR} = R - P = (1 + 2, 3 - 2, -4 - 4) = (3, 1, -8) \] Now, calculate the cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \): \[ \overrightarrow{n} = \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -7 \\ 3 & 1 & -8 \end{vmatrix} = \hat{i}((-2)(-8) - (-7)(1)) - \hat{j}(1(-8) - (-7)(3)) + \hat{k}(1(1) - (-2)(3)) \] \[ = \hat{i}(16 + 7) - \hat{j}(-8 + 21) + \hat{k}(1 + 6) = \hat{i}(23) - \hat{j}(13) + \hat{k}(7) \] Thus, the normal vector is \( \overrightarrow{n} = (23, -13, 7) \). 2. **Equation of the plane**: We use the point-normal form of the plane equation: \[ 23(x + 2) - 13(y - 2) + 7(z - 4) = 0 \] Expanding this: \[ 23x + 46 - 13y + 26 + 7z - 28 = 0 \] Simplifying: \[ 23x - 13y + 7z + 44 = 0 \] Rearranging gives us: \[ 23x - 13y + 7z + 44 = 0 \] Therefore, entering as \( A x + B y + C z \): \[ 23x - 13y + 7z + 44 = 0 \]