Answer
$x-2y+4z=0$
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:
$a(x-p)+b(y-q)+c(z-r)=0$
Here, the plane is parallel to the $x-2y+4z=6$ and having the normal vector $n=\lt 1,-2,4\gt$ can be expressed algebraically by the equation as follows:
$1(x-0)+(-2)(y-0)+4(z-0)=0$
or, $1(x-0)-2(y-0)+4(z-0)=0$
Thus, we have $x-2y+4z=0$
