Given three normalized vectors $$ \mathbf{u}=(\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3) $$ $$ \mathbf{v}=(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3) $$ $$ \mathbf{w}=(\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3) $$ $$ ||\mathbf{u}||=||\mathbf{v}||=||\mathbf{w}||=1 $$ What does

$$ \det\begin{vmatrix} \mathbf{u}_1 & \mathbf{u}_2& \mathbf{u}_3 \\ \mathbf{v}_1& \mathbf{v}_2& \mathbf{v}_3\\ \mathbf{w}_1& \mathbf{w}_2& \mathbf{w}_3 \end{vmatrix}=0 $$ mean (geometrical interpretation) ?

I also find out that $$ \det\begin{vmatrix} \mathbf{u}_1 & \mathbf{u}_2& \mathbf{u}_3 \\ \mathbf{v}_1& \mathbf{v}_2& \mathbf{v}_3\\ \mathbf{w}_1& \mathbf{w}_2& \mathbf{w}_3 \end{vmatrix}= \det\begin{vmatrix} \mathbf{u}_2 & \mathbf{u}_3& \mathbf{u}_1 \\ \mathbf{w}_2& \mathbf{w}_3& \mathbf{w}_1\\ \mathbf{v}_2& \mathbf{v}_3& \mathbf{v}_1\\ \end{vmatrix} $$ what's the geometrical implication of this equation?