Find the Equation of the Plane Perpendicular to Another Plane that Passes Through Origin

The question is as follows:

Write an equation for a plane that is perpendicular to the plane $2x − y + 3z = 6$ and that passes through the origin.

The normal vector I know would be $[a, b, c]$ where the variables belong in the equation $a(x_2 - x_1) + b(x_2 - x_1) + c(x_2 - x_1) = d$. Therefore, the normal vector of the given plane is $[2, -1, 3]$. Since it is going through the origin then the equation of the plane should be $2x - y + 3z = 0$. However, when I graph this online I get two planes that looks like they're parallel to each other.

Is there a place that I am making an error and, if so, where? Any help will be greatly appreciated.

• The plane that's perpendicular to the given plan will have normal vector perpendicular to the normal vector of the given plane. – Vasya Apr 7 '18 at 1:48

For example, as $(2,-1,3)\cdot(1,2,0)=0$, $x+2y=0$ is one of the possible answers.