Find the equation for the plane through the points \( P_{0}(0,5,-5), Q_{0}(3,-3,4) \), and \( R_{0}(-1,1,-4) \). The equation of the plane is \( \square \).

To find the equation of the plane through the points \( P_{0}(0,5,-5), Q_{0}(3,-3,4) \), and \( R_{0}(-1,1,-4) \), we need to follow these steps: 1. **Find two vectors in the plane** by using the given points: \[ \vec{PQ} = Q_{0} - P_{0} = (3-0, -3-5, 4-(-5)) = (3, -8, 9) \] \[ \vec{PR} = R_{0} - P_{0} = (-1-0, 1-5, -4-(-5)) = (-1, -4, 1) \] 2. **Calculate the normal vector** \( \vec{n} \) to the plane by taking the cross product of \( \vec{PQ} \) and \( \vec{PR} \): \[ \vec{n} = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -8 & 9 \\ -1 & -4 & 1 \end{vmatrix} \] This determinant results in: \[ \vec{n} = \hat{i}((-8)(1) - (9)(-4)) - \hat{j}((3)(1) - (9)(-1)) + \hat{k}((3)(-4) - (-8)(-1)) \] \[ = \hat{i}(-8 + 36) - \hat{j}(3 + 9) + \hat{k}(-12 - 8) \] \[ = \hat{i}(28) - \hat{j}(12) - \hat{k}(20) \] So, \( \vec{n} = (28, -12, -20) \). 3. **Use the point-normal form** of the equation of the plane \( n_{x}(x - x_0) + n_{y}(y - y_0) + n_{z}(z - z_0) = 0 \): Using point \( P_{0}(0,5,-5) \): \[ 28(x - 0) - 12(y - 5) - 20(z + 5) = 0 \] Expanding this gives: \[ 28x - 12y + 60 - 20z - 100 = 0 \] Simplifying further, we have: \[ 28x - 12y - 20z - 40 = 0 \] Thus, the equation of the plane is: \[ 28x - 12y - 20z = 40. \] So the final answer can be written as: \[ 28x - 12y - 20z = 40. \]