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