Are we to assume that "L"isthe line parallel to the given vector?

If so, you are misinterpreting the problem. The point (-2, 3, 1)cannotbe on the line. If it were, there would be an infinite number of planes passing through that line (and point). L must be anyotherline, parallel to that vector,notcontaining that point.

Write , , where you are free to choose as long as it gives a linenotpassing through (-2, 3, 1). For example, taking gives , , which does NOT pass through (-2, 3, 1) because if then .

You can find another vector in the plane by taking a vector from any given point on that line to (-2, 3, 1). For example, taking t= 0 gives (0, 0, 0) so another vector in the plane is <-2, 3, 1>. The cross product of that and the given vector is normal to the plane.