Final answer:
The scalar equation for the plane at a distance of 6 from the plane P with equation 2x+2y−z=3 is 2x + 2y - z - 3 = ±18.
Step-by-step explanation:
The plane that is at a distance of 6 from the plane P with equation 2x+2y−z=3 can be defined by a scalar equation. To find this equation, we can start by finding the equation of the plane P in normal form. The normal form of a plane equation is given by Ax + By + Cz = D, where (A, B, C) is the normal vector to the plane. In this case, the normal vector is (2, 2, -1) and D = 3. Now, the distance between a point (x, y, z) and the plane P can be found using the formula d = |Ax + By + Cz - D| / sqrt(A^2 + B^2 + C^2). We want this distance to be 6, so we can set up the equation |2x + 2y - z - 3| / sqrt(2^2 + 2^2 + (-1)^2) = 6 and solve for |2x + 2y - z - 3|. This gives us two possible equations: 2x + 2y - z - 3 = ±6 * sqrt(9), which simplifies to 2x + 2y - z - 3 = ±18. Therefore, the scalar equation for the plane at a distance of 6 from the plane P is 2x + 2y - z - 3 = ±18.