Solution: Let \[ \gamma (t) = \mathbf{A} t^2 + \mathbf{B} t + \mathbf{C}, ~t\in \mathbb{R}, ~\mathbf{A} ,\mathbf{B} ,\mathbf{C} \in \mathbb{R} ^3. \] Recall that a point $\mathbf{X} = (x,y,z)$ lies in a plane $P$ if \[ \left( \mathbf{X} -\mathbf{X}_0 \right) \cdot \mathbf{N} = 0, \] where $\mathbf{X} _0\in P$ and $\mathbf{N} $ is perpendicular to the plane. Equivalently, $\mathbf{X} \cdot \mathbf{N} = c $, where $c= \mathbf{X} _0\cdot \mathbf{N} $. Consider \[ \gamma (t) - \gamma (0) = \mathbf{A} t^2 + \mathbf{B} t + \mathbf{C} - \mathbf{C} = \mathbf{A} t^2 + \mathbf{B} t. \]

Note that \[ \left\{ \mathbf{A} t^2 + \mathbf{B} t: t \in \mathbb{R} \right\} \] is a $2$-dimensional vector space, so there exists $\mathbf{N} \in \mathbb{R} ^3$ such that $\left( \mathbf{A} t^2 + \mathbf{B} t \right) \cdot \mathbf{N} = 0$. So we got \[ \left( \gamma (t) - \gamma (0) \right) \cdot \mathbf{N} =0 \implies \gamma (t) \cdot \mathbf{N} = c. \] Hence, $\gamma (t)$ is contained in a plane.