Answer
$x=1,y=2+t, z=-t$
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 the vector $v=\lt 0,1,-1 \gt$ and $P=(1,2,0)$ .
Thus, we get the parametric equations:
$x=1+(0)t,y=2+t(1), z=0-t$
Hence, $x=1,y=2+t, z=-t$
