= Solution
A plane through the three points also contains the origin exactly when their <position vectors> are linearly dependent. The <scalar triple product> is
$$
\det\begin{pmatrix}1&2&3\\4&5&6\\\gamma&8&7\end{pmatrix}
=27-3\gamma.
$$
Thus $\boxed{\gamma=9}$. 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.
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