Let \(\vec{w}=\hat{i}+\hat{j}-2 \hat{k}\), and \(\overrightarrow{\mathrm{u}}\) and \(\overrightarrow{\mathrm{v}}\) be two vectors such that \(\vec{u} \times \vec{v}=\vec{w}\) and \(\vec{v} \times \vec{w}=\vec{u}\). Let \(\alpha, \beta, \gamma\) and \(t\) be real numbers such that \(\vec{u}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k},-t \alpha+\beta+\gamma=0, \alpha-t \beta+\gamma=0\), and \(\alpha+\beta-t \gamma=0\).
Match each entry in List-I to the correct entry in List-II and choose the correct option.

	LIST - I		LIST - II
(P)	\(|\vec{v}|^2\) is equal to	(1)	0
(Q)	If \(\alpha=\sqrt{3}\),then \(\gamma^2\) is equal to	(2)	1
(R)	If \(\alpha=\sqrt{3}\),then \((\beta+\gamma)^2\) is equal to	(3)	2
(S)	If \(\alpha=\sqrt{2}\),then \(t+3\) is equal to	(4)	3
		(5)	5

\(\begin{array}{lll}\vec{u} \times \vec{v}=\vec{w} & \vec{u} \perp \vec{v} & \vec{u} \perp \vec{w} \\ \vec{v} \times \vec{w}=\vec{u} & \vec{v} \perp \vec{w} & \end{array}\)
\(\begin{aligned} & \overrightarrow{\mathrm{w}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{u}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}\end{aligned} \quad\left|\begin{array}{ccc}-\mathrm{t} & 1 & 1 \\ 1 & -\mathrm{t} & 1 \\ 1 & 1 & -\mathrm{t}\end{array}\right|=0\)
\(\begin{array}{ll}\vec{u} . \vec{w}=0=\alpha+\beta-2 \gamma & t=-1 \text { or } 2 \\ (\vec{v} . \vec{w}) \times \vec{v}=\vec{u} \times \vec{v} & t=-1 \\ \vec{w}|\vec{v}|^2-0=\vec{w} & \alpha+\beta+\gamma=0 \\ |\vec{v}|=1(P) & t=2 \\ & \alpha=\beta=\gamma\end{array}\)
\(\begin{aligned} & \text { Also, } \vec{u} \times \vec{v}=\vec{w} \\ & |\vec{u}||\vec{v}|=|\vec{w}| \\ & \Rightarrow|\vec{u}|=|\vec{w}|=\sqrt{6}\end{aligned}\)
Case-1:
\(\begin{aligned}
& \mathrm{t}=2 \\
& \alpha=\beta=\gamma \\
& \alpha^2+\beta^2+\gamma^2=6 \\
& \alpha=\sqrt{2},-\sqrt{2} \\
& \text { (S) } \mathrm{t}+3=5
\end{aligned}\)
Case-2:
\(\begin{aligned}
& \mathrm{t}=-1 \\
& \alpha+\beta+\gamma=0 \\
& \alpha+\beta-2 \gamma=0 \\
& \gamma=0, \alpha=-\beta \\
& (\mathrm{Q}) \alpha=\sqrt{3}, \gamma^2=0, \beta=-\sqrt{3} \\
& (\mathrm{R}) \alpha=\sqrt{3}(\gamma+\beta)^2=\beta^2=3 \\
& (\mathrm{P}) \rightarrow(2),(\mathrm{Q}) \rightarrow(1),(\mathrm{R}) \rightarrow(4),(\mathrm{S}) \rightarrow(5)
\end{aligned}\)