Find the point of contact of the plane $2x-2y+z+12=0$ with the sphere $x^2+y^2+z^2-2x-4y+2z=3.$ 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 ?
 A: You can easily check that $(-3,4,0)$ is not on the plane given by $$2x-2y+z+12=0$$ because $$2(-3)-2(4)+0+12=-14+12=-2\ne 0$$. I don't understand what your solution is trying to do because you haven't explained what $u$, $v$ and $w$ are. Adding unknowns is usually not the way to go.
An easier way to solve this is to consider the wording. We are looking for a single point of contact which means that the plane is tangent to the sphere which means that the vector that is normal to the plane is parallel to the gradient of a function that has the sphere as a level surface.
Let $$f(x,y,z)=x^2+y^2+z^2-2x-4y+2z$$ Then the sphere is the set of points that satisfy $f(x,y,z)=3$ and its gradient will be normal to the sphere.
This gives $$(2x-2)\hat i +(2y-4)\hat j+(2z+2)\hat k=c(2\hat i-2\hat j+\hat k)$$ where $c$ is some constant. This gives three equations: $$2x-2=2c\implies x=1+c\\2y-4=-2c\implies y=2-c\\2z+2=c\implies z=\frac{c-2}{2}$$
We can plug these values back into the equation of the plane to get $c$.
$$2(1+c)-2(2-c)+(\frac{c-2}{2})+12=0\implies 4c+\frac{c-2}{2}+10=0\implies c=-2$$
Plugging this into our equations for $x$, $y$ and $z$ gives $$(x,y,z)=(-1,4,-2)$$
A: If you, as Alex K suggests above, allow for a scaling factor $k,$ $$(xx_1+yy_1+zz_1-(x+x_1)-2(y+y_1)+(z+z_1)-3)-(2x-2y+z+12)k=0$$
you get four equations
$$\begin{align}
x_1-2k-1&=0\\y_1+2k-2&=0\\z_1-k+1&=0\\-x_1-2y_1+z_1-12k-3&=0\end{align}$$
with solution $$k=-1,  x_1 =-1  y_1 = 4,z_1 =- 2.$$
A: If you know that $\langle A,B,C\rangle$ is a normal vector to the plane
$$A(x-x_1)+B(y-y_1)+C(z-z_1)=0  $$
then you may solve the exercise as follows.
Completing the square shows that the center of the sphere is $(1,2,-1)$ and the radius is $r=3$.
Since a normal vector (i.e. a vector orthogonal to the plane) is
$$ \langle 2,-2,1\rangle $$
the point of tangency will have the form
$$ (1,2,-1)+(2,-2,1)t=(1+2t,2-2t,-1+t) $$
Substituting into the equation of the plane gives
$$ 2(1+2t)-2(2-2t)+(-1+t+12=0 $$
Solving for $t$ gives $t=-1$.
Therefore
$$  (1+2t,2-2t,-1+t)=(-1,4,-2)$$
NOTE: In general, for a plane which does not cut a sphere, this approach gives either the tangent point or the point of the plane nearest the sphere. In this case you are given that the point is on the sphere, and that is easily checked to be the case.
