Question

Find an equation for the plane that is perpendicular to the line l(t) = (2, 0, 9)t + (3, −1, 1) and passes through (9, −1, 0).

Answer 1

The plane will have its normal vector equal to the direction vector of the line (2, 0, 9). Its equation is
  2x +9z = (2, 0, 9)•(9, -1, 0)
  2x +9z = 18