Answer
$\lt x,y,z \gt =\lt -3,0,2 \gt +t \lt 0,1,0 \gt$
Work Step by Step
The plane containing the point $P(p,q,r)$ and having the normal vector $n=\lt a,b,c\gt$ is expressed algebraically by the equation as follows:
$a(x-p)+b(y-q)+c(z-r)=0$
we are given that the line is perpendicular to the xz -plane, that is the line passes through $P(-3,0,2)$ and the line is parallel to $y$-axis ,then we have $n=\lt 0,1,0, \gt$
The line passing through passes through $P(-3,0,2)$ and parallel to the normal vector $n=\lt 0,1,0, \gt$ is expressed by the parametric equations as follows:
$x=-3; y=t; z=2$
Thus, the equation of the line is:
$\lt x,y,z \gt =\lt -3,0,2 \gt +t \lt 0,1,0 \gt$
