Sketch the plane $4x+2y+3z=24$. FInd the perpendicular distance of the plane $4x+2y=24$ from origin $O$ by first finding co-ordinates of the point $P$ on the plane such that $OP$ is perpendicular to the given plane.
What i tried: $OP=(4,2,3)$ thus length of $OP$ is $\sqrt{4^2+2^2+3^3}$ =$\sqrt{29}$ and the length of the plane is $\sqrt{4^2+2^2}$=$\sqrt{20}$
Thus by pythogoras theorem the perpendicular distane of the plane is $\sqrt{29-20}=3$
Thus the perpendicular distance of the plane $4x+2y=24$ from origin $O$ is 3 units. Am i correct?
Hint. A normal vector to a plane defined by $ax+by+cz=d$ is $(a,b,c)$.