15. Find a scalar equation of the plane that contains the origin and the point $(2,-3,2)$ and is perpendicular to the plane $x+2 y-z+3=0$ [5 marks]

Step 1 ：Find the normal vector of the given plane: \(\textbf{n}_{\text{given}} = (1, 2, -1)\)

Step 2 ：Find a vector in the plane containing the origin and the point (2, -3, 2): \(\textbf{v} = (2, -3, 2)\)

Step 3 ：Find the normal vector of the new plane by taking the cross product of the normal vector of the given plane and the vector in the plane: \(\textbf{n}_{\text{new}} = (1, -4, -7)\)

Step 4 ：Write the scalar equation of the plane using the normal vector and the fact that the plane contains the origin: \(\boxed{x - 4y - 7z = 0}\)