Find the point of contact of the plane $2x-2y+z+12=0$ with the sphere $x^2+y^2+z^2-2x-4y+2z=3.$
I already know the solution of this problem. However, while solving this, I first tried solving it in the following approach:
We know that the equation of a tangent plane of the sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$(,where $(-u,-v,-w)$ is the centre of the sphere) at the point $(x_1,y_1,z_1)$ is $xx_1+yy_1+zz_1+(x+x_1)u+(y+y_1)v+(z+z_1)w+d=0.$ Given, the equation of the plane $2x-2y+z+12=0$ and comparing it with $xx_1+yy_1+zz_1+(x+x_1)u+(y+y_1)v+(z+z_1)w+d=x(x_1+u)+y(y_1+v)+z(z_1+w)+x_1u+y_1v+z_1w+d=0,$ where $(u,v,w)=(-1,2,1)$ we get, $u+x_1=2,v+y_1=-2,w+z_1=1$ and hence, $x_1=3,y_1=0,z_1=0$ due to which $(3,0,0)$ is the point of contact.
Now, this solution is obviously wrong as the correct point of contact will be $(-1,4,-2).$ But I don’t understand what's going wrong with this approach? To be specific: I want to know, why we can't apply this approach ?