How to see a plane is tangent to a sphere from their equations Say you have two equations with three variables, the first is the equation  of the surface of a sphere and the second of a plane. In this case they intersect in a point $(1,0,0)$. The only way I know to find this point is to rewrite the equation of the sphere so you know its center point and intersect a line going through that point at and at an angle of 90 degrees with the plane. Are there other methods to solve this? Without geometry?
Here are two example equations.
$$\begin{cases}
x^2 + y^2 + z^2 - 6x + 6y - 12z + 5&=&0\\
2x - 3y + 6 z - 2&=&0\\
\end{cases}$$
and the solution
$$\begin{cases}
x = 1\\
y = 0\\
z = 0\\
\end{cases}$$
 A: First, a sphere and plane can intersect in a circle, a point, or not at all.  As  André Nicolas has said, $(1,0,0)$ does not satisfy your equations.  For your equations, Alpha shows that the intersection is a circle, not a point.
With the new equation, Alpha still thinks there are more than one point of intersection.  The first is now $(x-3)^2+(y+3)^2+(z-6)^2=7^2$ so $(1,0,0)$ is an intersection.
A: Let us solve the general case. Any sphere $S$ is characterized by an equation $s(x,y,z)=0$, where

$$
s(x,y,z)=x^2+y^2+z^2-2ax-2by-2cz+d,
$$

for some parameters $(a,b,c,d)$ such that $d\lt a^2+b^2+c^2$. Likewise, any plane $P$  is characterized by an equation $p(x,y,z)=0$, where

$$
p(x,y,z)=ux+vy+wz+t,
$$

for some parameters $(u,v,w,t)$ such that $(u,v,w)\ne(0,0,0)$. That $P$ is tangent to $S$ is equivalent to the condition that a point $(x,y,z)$ belongs to both $P$ and $S$, such that the line between the center $(a,b,c)$ of $S$ and the point $(x,y,z)$ is orthogonal to $P$.
The first part of the condition reads $p(x,y,z)=s(x,y,z)=0$. The vector $(u,v,w)$ is orthogonal to $P$ hence the second part of the condition is that $(x-a,y-b,z-c)$ and $(u,v,w)$ are proportional.
Thus $(x,y,z)=(a+\lambda u,b+\lambda v,c+\lambda w)$ for some $\lambda$. Then $p(x,y,z)=0$ if and only if
$$
(u^2+v^2+w^2)\lambda=ua+bv+cw+t,
$$
and $s(x,y,z)=0$ if and only if
$$
(u^2+v^2+w^2)\lambda^2=a^2+b^2+c^2-d.
$$
Thus the plane $P$ is tangent to the sphere $S$ if and only if these two equations have a common solution $\lambda$, which happens if and only if

$$
(ua+bv+cw+t)^2=(u^2+v^2+w^2)\cdot(a^2+b^2+c^2-d).
$$

A: A simple approach is to realize that, for a plane PL to be tangent to a sphere SPH, then the distance of the sphere center SPH.c from the plane PL has to be SPH.r.
Suggestion: If you represent the plane using [ orientation, location ] numerics, then it's straightforward to calculate the distance of arbitrary point p from the plane:
distance <-- abs ( p • orientation  - location )
The orientation of the plane is its normalized normal vector.
The location of the plane is its invariant coordinate along the axis defined by orientation.  If you know a point p0 on the plane, 
location <-- p0 • orientation
If you're starting with an algebraic equation form
ax + by + cz = d
divide both sides of the equation by sqrt (a^^2 + b^^2 + c^^2).  Then, rewrite as the vector dot product equation:
p • o = L
where p is any point on the plane p = ( x y z )
  o is the orientation of the plane o = < x y z >

  and L is the plane Location coordinate (signed distance from the origin)

