Answer
$\lt x,y,z \gt =\lt -2,0,0 \gt -t \lt -2,0,-10 \gt$
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 passing through x-axis ,so we have $x=-2,y=0,z=0$ .This means that $P(-2.,0,0)$
Also, on the $z$-axis ,then we have $x=0,y=0,z=10$, that is, $Q(0,0,10)$
The line passing through the x-axis point $P(-2,0,0)$ and $Q(0,0,10)$and parallel to the normal vector $n=\lt a,b,c\gt$ is expressed by the parametric equations as follows:
$x=p+at; y=q+bt; z=r+ct$
Then, $n=P(-2,0,0) -Q(0,0,10)=\lt -2,0,-10 \gt$
Thus, the parametric equations are:
$x=-2-2t; y=0; z=-10t$
The equation of the line is:
$\lt x,y,z \gt =\lt -2,0,0 \gt -t \lt -2,0,-10 \gt$
