Chapter 14 - Calculus of Vector-Valued Functions - 14.4 Curvature - Preliminary Questions - Page 734: 4

$$\kappa(s)= 0$$

Since $ r(t) = \lt2 + 3t, 7t, 5 − t\gt$, then $ r'(t) = \lt 3, 7, -1\gt$ and hence $\|r'(t)\|=\sqrt{9+49+1}=\sqrt{59}$, $T(t)=\frac{r'(t)}{\|r'(t)\|}=\frac{ \lt 3, 7, -1\gt}{\sqrt{59}}$. Now, the curvature is given by $$\kappa(s)=\frac{1}{\|r'(t)\|}\|\frac{dT}{dt}\|=0$$ since the tangent vector $T$ is constant.