$ \vec{u} \times \vec{v}=\vec{w} \text { and } \vec{v} \times \vec{w}=\vec{u} $ $\Rightarrow \quad \vec{u}, \vec{v}$ and $\vec{w}$ are mutually perpendicular. $ \begin{aligned} & (\vec{v} \times \vec{w}) \times \vec{v}=\vec{w} \\ & \Rightarrow \quad \vec{w}(\vec{v} \cdot \vec{v})-\vec{v}(\vec{v} \cdot \vec{w})=\vec{w} \\ & \Rightarrow \quad \vec{w}\left(|\vec{v}|^2-1\right)-\vec{v}(\vec{v} \cdot \vec{w})=\overrightarrow{0} \\ & \Rightarrow \quad|\vec{v}|^2=1 \text { and } \vec{v} \cdot \vec{w}=0 \\ & |\vec{u}||\vec{v}|=|\omega| \Rightarrow|\vec{u}|=\sqrt{6} \\ & \vec{u} \cdot \vec{w}=0 \Rightarrow \alpha+\beta-2 \gamma=0 \\ & \text { and }-t \alpha+\beta+\gamma=0 \\ & \alpha-t \beta+\gamma=0 \\ & \alpha+\beta-t \gamma=0 \end{aligned} $ $ \begin{aligned} & (i i)-(i) \Rightarrow \alpha(1+t)=(t+1) \beta \\ & (i i i)-(i i) \Rightarrow \beta(1+t)=(1+t) \gamma \\ & \Rightarrow \alpha(1+t)=\beta(1+t)=\gamma(1+t) \end{aligned} $ either $t=-1$ or $\alpha=\beta=\gamma$ $ \begin{aligned} & \Rightarrow \quad \sqrt{\alpha^2+\alpha^2+\alpha^2}=\sqrt{6} \\ & \Rightarrow \quad \alpha=\sqrt{2},-t \alpha=-\alpha-\alpha \\ & \Rightarrow \quad t=2 \end{aligned} $ Since $\alpha=\sqrt{3}$ $ \begin{aligned} \Rightarrow & t=-1 \\ \Rightarrow & \alpha+\beta+\gamma=0 \\ & \alpha+\beta-2 \gamma=0 \\ \Rightarrow & \gamma=0 \end{aligned} $ (Q) $\rightarrow 1$ $ \begin{aligned} & \alpha+\beta=0 \Rightarrow(\beta+\gamma)=-\alpha \Rightarrow(\beta+\gamma)^2=\alpha^2=3,(R) \rightarrow 4 \\ & \Rightarrow|\vec{v}|^2=1 \rightarrow(P) \rightarrow 2 \end{aligned} $ If $\alpha=\sqrt{3} \Rightarrow \gamma^2=0 \rightarrow(Q) \rightarrow 1$ If $\alpha=\sqrt{3} \Rightarrow(\beta+\gamma)^2=(\sqrt{3})^2=3,(R) \rightarrow 4$ If $\alpha=\sqrt{2}, t+3=(2)+3=5,(S) \rightarrow 5$