I would agree that you've found two points on the line. Now what should you do?
ok, let me start again i have the three points (6,6,-1) (13,10,-2) (1,3,2)
$\displaystyle n=OB-OA x OC-OA$
$\displaystyle n= 7i+4j-k x -5i-3j+3k$
find the cross product
$\displaystyle i(12+3)-j(21-5)+k(-21+20)$
$\displaystyle 15i-16j-k$
sub points (1,3,2)
i get $\displaystyle 15(1)-16(3)-(2)$
$\displaystyle 15-48+2=-35$
$\displaystyle 15x-16y-z=-35$
this is incorrect, however shouldn't it be still correct even if the points are different
Name the three points: $\displaystyle P6,6,-1),~Q13,10,-2),~\&~R1,2,3)$.
Now find the normal: $\displaystyle N = \overrightarrow {PQ} \times \overrightarrow {PR} $.
Then write the equation of the plane determined by the three points:
$\displaystyle N \cdot < x - 1,y - 2,z - 3 > = 0$
BTW: [tex]\times [/tex] gives $\displaystyle \times $ in LaTeX.