Points A (2; m; -1) and B (1; 2; m) and plane 2х-3у + z-1 = 0 are given. At what values of m is this plane

Points A (2; m; -1) and B (1; 2; m) and plane 2х-3у + z-1 = 0 are given. At what values of m is this plane parallel to the straight line AB?

Let us find the coordinates of the vector AB: ((1 – 2); (2 – m); (m – (-1)) = (-1; 2 – m; m + 1). From the plane equation we obtain the coordinates of the direction vector: (2 ; -3; 1). Since the plane and the line are parallel, the scalar product of AB and the direction vector is 0. We obtain the equation:

-1 * 2 – 3 * (2 – m) + (m + 1) = 0;

-2 – 6 + 2m + m + 1 = 0;

3m = 7;

m = 7/3.