Let vec{w}=hat{i}+hat{j}-2 hat{k},and vec{u} | vedclass

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(A) Given $\vec{w} = \hat{i} + \hat{j} - 2\hat{k}$. Since $\vec{u} 	imes \vec{v} = \vec{w}$,we have $\vec{u} \perp \vec{w}$ and $\vec{v} \perp \vec{w

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$(P) \rightarrow (2), (Q) \rightarrow (1), (R) \rightarrow (4), (S) \rightarrow (5)$

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(A) Given $\vec{w} = \hat{i} + \hat{j} - 2\hat{k}$. Since $\vec{u} 	imes \vec{v} = \vec{w}$,we have $\vec{u} \perp \vec{w}$ and $\vec{v} \perp \vec{w}$.Also,$\vec{v} 	imes \vec{w} = \vec{u}$,which implies $\vec{u} \perp \vec{w}$ and $\vec{v} \perp \vec{w}$.The system of equations $-t\alpha + \beta + \gamma = 0$,$\alpha - t\beta + \gamma = 0$,$\alpha + \beta - t\gamma = 0$ has a non-trivial solution if the determinant is zero:$\begin{vmatrix} -t & 1 & 1 \ 1 & -t & 1 \ 1 & 1 & -t \end{vmatrix} = 0 \Rightarrow -(t^3 - 1) - 1(-t - 1) + 1(1 + t) = 0 \Rightarrow -t^3 + 1 + t + 1 + 1 + t = 0 \Rightarrow t^3 - 2t - 3 = 0$.Factoring gives $(t+1)(t^2 - t - 3) = 0$. For the given conditions,$t = -1$ or $t = 2$.If $t = 2$,$\alpha = \beta = \gamma$. Since $\vec{u} \cdot \vec{w} = 0$,$\alpha + \beta - 2\gamma = 0 \Rightarrow 2\alpha - 2\alpha = 0$,which is consistent.Using $|\vec{u}| = |\vec{w}| = \sqrt{6}$,$3\alpha^2 = 6 \Rightarrow \alpha^2 = 2$. Thus $\alpha = \pm \sqrt{2}$. For $t=2$,$t+3 = 5$.If $t = -1$,$\alpha + \beta + \gamma = 0$. Also $\vec{u} \cdot \vec{w} = 0 \Rightarrow \alpha + \beta - 2\gamma = 0$.Subtracting gives $3\gamma = 0 \Rightarrow \gamma = 0$. Then $\beta = -\alpha$.For $\alpha = \sqrt{3}$,$\gamma^2 = 0$ and $(\beta + \gamma)^2 = (-\sqrt{3} + 0)^2 = 3$.Thus,$(P) ightarrow (2), (Q) ightarrow (1), (R) ightarrow (4), (S) ightarrow (5)$.

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	$(4)$ $3$
	$(5)$ $5$

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