If [tex]P=(x,y,z)[/tex] is any point in the plane and [tex]P_0=(6,0,4)[/tex] is the given point, then [tex](x-6)\,\mathbf i+y\,\mathbf j+(z-4)\,\mathbf k[/tex] will be a vector that lies in the plane.
Now, if [tex]2\,\mathbf i-7\,\mathbf j+3\,\mathbf k[/tex] is normal to the plane, then the plane is given by the equation
[tex](2\,\mathbf i-7\,\mathbf j+3\,\mathbf k)\cdot((x-6)\,\mathbf i+y\,\mathbf j+(z-4)\,\mathbf k)=0[/tex]
[tex]2(x-6)-7y+3(z-4)=0[/tex]
[tex]2x-7y+3z=24[/tex]
