The equation of a sphere tangent to a plane at a point and also tangent to another plane

Question

Determine the equation of the sphere that is tangent to the plane $$(P):6x-3y-2z-35=0$$ at $M(5,-1,-1)$ and also tangent to the plane $$(Q):6x-3y-2z+63=0$$

My solution: Let $d$ be the line passing through $M$, $d$ orthogonal to $(P)$. The parametric equations of the line $d$ are: $$x = 5+6t$$ $$y=-1-3t$$ $$z=-1-2t$$ So the center of the sphere is of the form $C(5+6t, -1-3t,-1-2t)$. The center of the sphere must belong to the line $d$ and the plane $(Q)$, so it is the intersection point between the line and the plane. $$6(5+6t)+3(1+3t)+2(1+2t)+63=0$$ From this relation we get $t=2$, so the center has the coordinates $C(17,-7,-5)$. Now since it's stated that the plane $(Q)$ is tangent to the sphere, the distance from the center to the plane is equal with the radius. The radius $r=14$, so now we have the equation of the sphere: $$(x-17)^2+(y+7)^2+(z+5)^2=14^2$$ Is this correct? I'm not sure if my initial condition for the center to belong to the line $d$ and the plane $(Q)$ is right, but I based my assumption off of the fact that they are both tangent to the sphere and so the distance from the center to each plane is equal with the radius. EDIT: Okay, so my assumption was definitely wrong, I somehow didn't notice the normal vectors were identical

Answer 1

No, that is not correct. You already know that the center of the sphere is of the form $(5+6t,-1-3t,-1-2t)$, for some real $t$. Since the planes $P$ and $Q$ and parallel, $C$ has to be the midpoint between $M$ and the point $M'$ which belongs to the line $(5+6t,-1-3t,-1-2t)$ and to $Q$. But $M'=(-7,5,3)$, and therefore $C=(-1,2,1)$ (and the radius is $7$).

Answer 2

Hint

Your sentence

The center of the sphere must belong to the line $d$ and the plane $(Q)$...

is not correct. As the sphere is supposed to be tangent to the plane $(Q)$ its center does not lie in $(Q)$.

However, you can use the fact that the center of the sphere lies on the line $d$ and that its distance to the plane $(P)$ should be equal to its distance to the plane $(Q)$.

Answer 3

(New user here, so I don't have enough reputation to comment)

Why should $Q$ contain the center of the sphere? In fact, if it does, is $Q$ still tangential?

Here's some motivation for a possible solution: Start with the line $d$. Since the two given planes are parallel, can you see why $d$ must contain the center of the sphere? Now try to use this fact, along with the distance between the planes (which is standard computation) to figure out the center/radius of the sphere.