## Calculus (3rd Edition)

$${\bf{r}}(t)= \langle 1+2t,1-6t,1+t \rangle.$$
To find a vector parametrization for the given line, we first find the direction vector: $(3,-5,2)-(1,1,1)=\langle 2,-6,1 \rangle$ Hence the parametrization is given by $${\bf{r}}(t)= \langle 1,1,1 \rangle +t \langle 2,-6,1 \rangle\\ = \langle 1+2t,1-6t,1+t \rangle.$$