Chapter 15 - Section 15.1 - Line Integrals - Exercises - Page 826: 6

$\bf {Graph(b)}$

Since, we have $r(t)=tj+(2-2t)k$ or, $x=0; y=t,z=2-2t$ This shows a line $z=2-2y$ along the plane $x=0$ whose intercept form is $y+\dfrac{z}{2}=1$. So, we have: $y$ intercept $=1$ and z-intercept $=2$ Hence, the required answer is $\bf {Graph(b)}$