Chapter 12 - Vector-Valued Functions - 12.1 Introduction To Vector-Valued Functions - Exercises Set 12.1 - Page 846: 40

See proof

Let us put $x = t,$ $y = \frac{1 + t}{t},$ $z = \frac{1 - t^2}{t}.$ Then $x - y + z + 1 = t - \frac{1 + t}{t} + \frac{1 - t^2}{t} + 1 = \frac{t^2 - 1 - t + 1 - t^2 + t}{t} = 0.$ Thus, any point of the parametric curve lies on the given plane, i.e., the entire curve lies on the plane.