Find the equation of the plane that satisfies the stated conditions.

The plane that contains the line x = -2 + 3t, y = 4 + 2t, z = 3 - t and is perpendicular to the plane x - 2y + z = 5.
 
Nevermind, I kinda just solved it myself.

All good.
 
Find the equation of the plane (if you dare) that is tangent to the following spheres:

sphere 1: r=2 center P(2,2,2)

sphere 2: r=3 center Q(3,4,5)

Ok, I know that the plane will be parallel to the vector PQ = <1,2,3>

For the equation of a plane I need a normal vector, and a point on the plane. I know that one of the points on the plane will be one of the points on the sphere (either sphere will do, right?).

The equations for the spheres are:

(x-2)^2 + (y-2)^2 + (z-2)^2=4

and

(x-3)^2 + (y-4)^2 + (z-5)^2=9

What i am thinking now is that i should take the midpoint between the spheres' centers. Use that to find another vector parallel to the plane in question, and then generate a normal vector from that - is this correct thinking?

And from there, i am unsure how to find a point common to the plane and the sphere (either sphere).

I have drawn this out on paper and understand the geometry...
 

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