Linear Algebra - Finding the equation of a plane from 3 points

[cris623]

Homework Statement

Find the equation of the plane which contains the points (−3 −1 3), (−5 −4 2) and (−6 0 0).
Write the equation in the form Ax+By+Cz=D

Homework Equations

none

The Attempt at a Solution

P (-3 -1 3)
Q (-5 -4 2)
R (-6 0 0)

Alright so first i found the vectors PQ and PR to be (-2 -3 -1) and (-3 1 -3)
Then I found the cross product PQ x PR to be (8 -3 -11) and put that into the equation to get 8x-3y-11z=D

and then subbed point P into that equation to solve for D and finished with:
8x-3y-11z=-54

This answer was marked wrong, I've checked everything over and over and can't figure out how to do it.

 

[Unit]

Check your cross product! I got (-2, -3, -1) cross (-3, 1, -3) = (10, -3, -11)

 

[cris623]

I'm still new with cross products and I swear I checked it like 10 times. I messed up when subtracting (-1) from 9, instead I subtracted (+1) from 9 to get 8 rather than 10. Geeze.

Thanks a lot!