Find the equation of the plane containing the point \( (-1,-4,-8) \) and parallel to the plane \( 7 x-6 y-7 z=0 \).

To find the equation of the plane that contains the point \((-1, -4, -8)\) and is parallel to the plane \(7x - 6y - 7z = 0\), follow these steps: 1. **Identify the Normal Vector:** - The given plane \(7x - 6y - 7z = 0\) has a normal vector \(\mathbf{n} = (7, -6, -7)\). - Since the desired plane is parallel to this one, it shares the same normal vector. 2. **Use the Point-Normal Form:** - The general equation of a plane with normal vector \((A, B, C)\) that passes through a point \((x_0, y_0, z_0)\) is: \[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \] - Plugging in the normal vector \((7, -6, -7)\) and the point \((-1, -4, -8)\): \[ 7(x + 1) - 6(y + 4) - 7(z + 8) = 0 \] 3. **Simplify the Equation:** - Expanding the equation: \[ 7x + 7 - 6y - 24 - 7z - 56 = 0 \] - Combine like terms: \[ 7x - 6y - 7z - 73 = 0 \] - Rearranging gives: \[ 7x - 6y - 7z = 73 \] **Final Equation of the Plane:** \[ 7x - 6y - 7z = 73 \] Answer: After simplifying, an equation for the plane is 7 x – 6 y – 7 z = 73