I got a maths question that gives you 3 points, $A (6,0,0)$ and $B (6,6,0)$ and $C (0,6,0)$ and a line DG, D being $(0,0,6)$ and $G (0,6,6)$ so the equation of DG is $\vec r$ = 6$\hat i$ -$6t\hat j$ . You're then asked to find the equation of the sphere that contains those 3 points and is tangential to the line.
I know that if I had 4 points I could do a simultaneous equation(s), but I don't see how that could work here. Please help.
I should maybe mention that the $origin, A, B, C, D, G, E(6,0,6) ~and~ F (6,6,6)$ form the 3 corners of a cube.
A sphere can be represented as $$(x-a)^2+(y-b)^2+(z-c)^2=r^2$$ where the center is $(a,b,c)$ and its radius is $r$.
Setting each of $(x,y,z)=(6,0,0),(6,6,0),(0,6,0)$ in the equation, you'll get $$(6-a)^2+(0-b)^2+(0-c)^2=r^2,$$ $$(6-a)^2+(6-b)^2+(0-c)^2=r^.2$$ $$(0-a)^2+(6-b)^2+(0-c)^2=r^2.$$
Hence, you'll get $$(a,b,c)=(3,3,r^2-18).$$
Now we know that the equation of the sphere is represented as $$(x-3)^2+(y-3)^2+(z-r^2+18)^2=r^2.$$
Then, you can set $$(x,y,z)=(0, 6t, 6)$$ in the equation (because the line $DG$ is represented as $(0,0,6)+t(0,6,0)$.)
Then, you'll have an equation with $t,r$, so what you do is to find $r$ such that the equation has only one real solution $t$.