The question is as follows:
Show that the sphere $(x−5)^2 +(y+2)^2 +(z−4)^2 = 36$ and the plane $2x+y+2z = 34$ have exactly one point in common.
The problem gave the hint that solving the two equations simultaneously will not be a good approach to this problem. If they do not want for me to subsitute one equation into the other, then how will I be able to solve this problem? Do they want parametric equations? Any help will be greatly appreciated.
Having only one point in common occurs if and only if the plane is tangent to the sphere, so we can try to show the plane is a tangent plane. This will occur if the minimum distance from the plane to the centre of the sphere is the radius of the sphere, i.e. $6$. The minimum distance occurs at a point where the normal line passes through the centre of the sphere, essentially by Pythagoras. The unit normal to the plane is $(2,1,2)/3$, and the point on the sphere in this direction from the centre is $$ (5,-2,4)+6\left( \frac{2}{3},\frac{1}{3},\frac{2}{3} \right) = (5,-2,4)+(4,2,4) = (9,0,8). $$ Then $(9,0,8) \cdot (2,1,2) = 18+16=34$, so $(9,0,8)$ indeed lies in the plane as well. Hence it is the only point in common.