Answer
$x=1+t; y=1-t; z=1+t$
Work Step by Step
The line passing through containing the point $P(p,q,r)$ and parallel to the normal vector $n=\lt a,b,c\gt$ is expressed by the parametric equations as follows:
$x=p+at; y=q+bt; z=r+ct$
Here, $t$ is any real number.
Now, the line passing through containing the point $P(1,1,1)$ and parallel to the normal vector $n=i-j+k=\lt 1,-1,1 \gt $ is expressed by the parametric equations as follows:
$x=1+t; y=1-t; z=1+t$
