Final answer:
To find the equation of the plane through a given point and perpendicular to the intersection line of two planes, calculate the cross product of the normal vectors of the given planes to get the normal of the required plane. Then, use the point-normal form of the plane equation with the given point and the calculated normal vector to find the final equation of the plane.
Step-by-step explanation:
To find a plane through P0 (2, 1, -1) and perpendicular to the line of intersection of the planes 2x + y - z = 3 and x + 2y + z = 2, we must first determine the normal vector to the line of intersection of the given planes. To do this, we calculate the cross product of the normal vectors of each plane.
The normal vector to the plane 2x + y - z = 3 is (2, 1, -1), and the normal vector to the plane x + 2y + z = 2 is (1, 2, 1). The cross product of these two vectors is (3, -4, 3), which is the normal vector to the plane we're trying to find.
Finally, the equation of the plane through P0 perpendicular to the line of intersection can be found using the point-normal form of the plane equation: A(x - x0) + B(y - y0) + C(z - z0) = 0, where (A, B, C) is the normal vector and (x0, y0, z0) is the point P0. Plugging in the values, we get 3(x - 2) - 4(y - 1) + 3(z + 1) = 0, which simplifies to 3x - 4y + 3z - 2 = 0 as the plane equation.