Intersection point of all tangent planes to a sphere with point of contact on a circle I have solved the following problem, but I am not sure about the way I have solved the third question, so I would be grateful if someone would check it out.
"Consider the plane $\pi: 2x+y+2z+6=0$ and the sphere $S$ with centre $C(-2,1,-1)$ and radius $3$.
(1) Find the centre and the radius of the circle $K$ intersection of the sphere and the plane;
(2) Let $J$ be the point $(0,0,-3)$. Show that $J$ lies on $S$ and find the equation of the plane tangent to $S$ at $J$.
(3) All the tangent planes to the sphere $S$ whose point of contact lie on the circle $K$ intersect at a point $Q$. Find the coordinates of $Q$.
My solution:
(1) $$d=\text{distance center of sphere-plane}=\frac{|ax_c+by_c+cz_c+d|}{\sqrt{a^2+b^2+c^2}}=\frac{|2(-2)+1\cdot 1+2\cdot (-1)+6|}{\sqrt{2^2+1^2+2^2}}=\frac{1}{3}.$$
Radius $$r=\sqrt{R^2-d^2}=\sqrt{9-\frac{1}{9}}=\fbox{$\sqrt{\frac{80}{9}}$}$$
Center $$C=(-2,1,-1)+d\frac{\vec{n}}{|\vec{n}|}=(-2,1,-1)+\frac{1}{3}\cdot\frac{(2,1,2)}{3}=\fbox{$(-\frac{16}{9},\frac{10}{9},-\frac{7}{9})$}.$$
(2) The equation of the sphere $S$ is $(x+2)^2+(y-1)^2+(z+1)^2=9$.
$J\in S$ since it satisfies this equation.
The equation of the tangent plane at $J$ is $\big( 2(x+2),2(y-1),2(z+1) \big)_{|(0,0,-3)}\cdot \big( (x-0),(y-0),(z+3) \big)=0\Leftrightarrow \fbox{$4x-2y-4z=12$}$
(3) Intuitively I would expect that these tangent planes form a kind of cone and they all meet at the point where each of them intersect the line passing through the center of the sphere $S$ and the circle $K$; such a line has equation $\vec{r}(t)=(-2,1,-1)+t(2,1,2)$ and its intersection with the plane found in (2) is $\fbox{$(-20,-8,-19)$}$, which I would assume belongs to each of the planes tangent to $S$ at a point in $K$ and is thus $Q$. Is this correct? Is there a way to do this rigorously? Thanks.
 A: There is a mistake in your work for the first question.
For the first question you assumed the center in the opposite direction. It is in fact
$C = (-2,1,-1) - d\frac{\vec{n}}{|\vec{n}|} =(-2,1,-1) -\frac{1}{3}\cdot\frac{(2,1,2)}{3}$
$=\fbox{$\left(-\frac{20}{9},\frac{8}{9},-\frac{11}{9}\right)$}$
Here is how I would do the first question. Say the center of the circle is $(x_0, y_0, z_0)$. Then,
$(x_0 + 2, y_0 - 1, z_0 + 1) = \lambda (2, 1, 2)$
$x_0 = - 2 + 2 \lambda, y_0 = 1 +  \lambda, z_0 = - 1 + 2 \lambda$
As the center of the circle is in the plane $2x + y + 2z + 6 = 0$, we get $\lambda = - \frac 19$.
Once we find the center, the radius is given by
$r^2 = 3^2 - 9 \lambda^2 \implies r = \frac{\sqrt{80}}{3} $
For the second question, your work is correct. Alternatively, the vector from center of the sphere to point $(0, 0, -3)$ is normal to the tangent plane so equation of the tangent plane is $2x - y - 2 (z + 3) = 0$
For the third question, your idea and work is correct. You need to make a note that as $(0, 0, -3)$ is on the sphere and also on the plane $2x + y + 2z + 6 = 0$, it is on the circle $K$. Tangent planes to the sphere at points that are on the circle $K$ would meet at the line through the center of the circle and orthogonal to the plane that the circle is contained in.
