Find an equation of the tangent plane to the surface at the given point. \[ z=8-\frac{8}{3} x-y, \quad(3,-5,5) \]

To find the equation of the tangent plane to the surface defined by \( z = 8 - \frac{8}{3} x - y \) at the point \( (3, -5, 5) \), we first need to determine the gradient or the partial derivatives of the function with respect to \( x \) and \( y \). 1. The surface can be rewritten as \( F(x, y, z) = z + \frac{8}{3} x + y - 8 = 0 \). 2. Now, calculating the partial derivatives: - \( F_x = \frac{\partial F}{\partial x} = \frac{8}{3} \) - \( F_y = \frac{\partial F}{\partial y} = 1 \) - \( F_z = \frac{\partial F}{\partial z} = 1 \) 3. The gradient \( \nabla F \) at the point \( (3, -5, 5) \) is \( \left( \frac{8}{3}, 1, -1 \right) \). 4. Using the point-normal form of the plane equation: \[ F_x (x - x_0) + F_y (y - y_0) + F_z (z - z_0) = 0 \] substituting \( (x_0, y_0, z_0) = (3, -5, 5) \): \[ \frac{8}{3}(x - 3) + 1(y + 5) - 1(z - 5) = 0 \] 5. Simplifying this equation, we get: \[ \frac{8}{3}x - \frac{8}{3} \cdot 3 + y + 5 - z + 5 = 0 \] 6. Which leads to: \[ \frac{8}{3}x + y - z + \frac{6}{3} = 0 \] Thus, the equation of the tangent plane at the point \( (3, -5, 5) \) is: \[ \frac{8}{3}x + y - z + 2 = 0 \]