Answer
$x=3+2t,y=-2-t, z=1+3t$
Work Step by Step
The parametric equations of a straight line can be found by knowing the value of a vector, such as $v=v_1i+v_2j+v_3k$, passing through a point $P(x_0,y_0,z_0)$ as follows:
$x=x_0+t v_1,y=y_0+t v_2; z=z_0+t v_3$
Here, we have $P(3,-2,1)$ and $v=\lt 2,-1,3 \gt$
Thus, we get the parametric equations:
$x=3+2t,y=-2-t, z=1+3t$
