Answer
$x=1+3t,y=2t,z= -1+t$
Work Step by Step
Given: $x=4+3t,y=2t,z= -2+t$
The direction of the vectors is $\lt3,2,1 \gt$
$r=r_{0}+tv$
where $r_{0}$ is the starting point vector and $v$ is the direction vector.
$r=\lt 1,0,-1\gt+t \lt 3,2,1\gt$
Therefore, the parametric equations of the line are:
$x=1+3t,y=2t,z= -1+t$
