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.

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 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...

 Tags calculus, multivariable