Answer
$\lt x,y,z \gt =\lt 0,0,4 \gt -t \lt -2,-5,4 \gt$
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 containing the point on the z-axis is $x=0,y=0,z=4$ .This means that $P(0,0,4)$
Also, on the $xy$-palne ,we have $x=2,y=5,z=0$, that is, $Q(2,5,0)$
The line passing through containing the point $P(0,0,4)$ and $Q(2,5,0)$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, $n=P(0,0,4)-Q(2,5,0)=\lt -2,-5,4 \gt$
Thus, the parametric equations are:
$x=-2t; y=-5t; z=4+4t$
The equation of the line is:
$\lt x,y,z \gt =\lt 0,0,4 \gt -t \lt -2,-5,4 \gt$
