 Solution

A plane through the three points also contains the origin exactly when their position vectors are linearly dependent. The scalar triple product isThus $\boxed{\gamma =9}$. The resulting plane is $\boxed{x-2y+z=0}$, which contains all three points and the origin. The points are not collinear: the differences $(3,3,3)$ and $(\gamma -1,6,4)$ cannot be proportional, since their last two components would require $6=4$. Therefore the plane used in the argument is well defined.