Equation of Plane w/ Origin 3 Units Away: Solved

[nhartung]

Homework Statement

Find the equation of a plane with distance 3 units from the origin and perpendicular to the line through P(1,2,3) and Q(-2,4,1).

Homework Equations

\vec{n} = \frac{\vec{PQ}}\left|{\vec{PQ}}\left|

Plane Equation: a(x-x₀) + b(y-y₀) + c(z-z₀)

The Attempt at a Solution

Ok so I think I can solve this all the way up until the end.

We have \vec{PQ} = <-3, -2, 2>

so \vec{n} = \frac{1}{\sqrt{17}}<-3,-2,2>

Now if I scale that vector by 3 or -3 I can get a point on the plane that I am looking for, I need to put this into the form of a plane so I use the equation above and end up getting:
\frac{-3}{\sqrt{17}}x + \frac{2}{\sqrt{17}}y - \frac{2}{\sqrt{}17}z = \frac{-27}{17} + \frac{18}{17} + \frac{18}{17}

This can be simplified further but this is where mine and my professors work differs. He gets the following on the right side of the equation = \frac{27}{17} + \frac{12}{17} + \frac{12}{17}

Any ideas?

 

[nhartung]

nevermind I figured it out.