#### Answer

See proof below.

#### Work Step by Step

$({\bf u}\times{\bf v})\times{\bf w}$ is perpendicular to both $({\bf u}\times{\bf v})$ and ${\bf w}$.
${\bf u}\times{\bf v}$ is perpendicular to both ${\bf u}$ and ${\bf v}$, if neither of them are zero vectors and they are not parallel (which is the degenerate case in which ${\bf u}\times{\bf v}$=${\bf 0}$).
So $({\bf u}\times{\bf v})\times{\bf w}$ is parallel to some vector in the plane determined by ${\bf u}$ and ${\bf v}$.
Therefore, $({\bf u}\times{\bf v})\times{\bf w}$ lies in the plane determined by ${\bf u}$ and ${\bf v}$.
${\bf u}\times({\bf v}\times{\bf w})$ is perpendicular to both ${\bf u}$ and $({\bf v}\times{\bf w})$.
$({\bf v}\times{\bf w})$ is perpendicular to both ${\bf v}$ and ${\bf w}$, if neither of them are zero vectors and they are not parallel (which is the degenerate case in which ${\bf v}\times{\bf w}$=${\bf 0}$).
So ${\bf u}\times({\bf v}\times{\bf w})$ is parallel to some vector in the plane determined by ${\bf v}$ and ${\bf w}$.
Therefore, ${\bf u}\times({\bf v}\times{\bf w})$ lies in the plane determined by ${\bf v}$ and ${\bf w}$.