Question 16

1. From $\vec{u} \times \vec{v} = \vec{w}$ and $\vec{v} \times \vec{w} = \vec{u}$, it follows that $\vec{u}, \vec{v}, \vec{w}$ are mutually orthogonal vectors. Taking magnitudes, we have $|\vec{u}||\vec{v}| = |\vec{w}|$ and $|\vec{v}||\vec{w}| = |\vec{u}|$. This implies $|\vec{v}|^2 = 1$ and $|\vec{u}| = |\vec{w}|$. Since $|\vec{w}|^2 = 1^2 + 1^2 + (-2)^2 = 6$, we have $|\vec{u}|^2 = \alpha^2 + \beta^2 + \gamma^2 = 6$.

2. The system of equations for $\alpha, \beta, \gamma$ is: $-t\alpha + \beta + \gamma = 0$ $\alpha - t\beta + \gamma = 0$ $\alpha + \beta - t\gamma = 0$ For non-trivial solutions, the determinant must be zero: $\begin{vmatrix} -t & 1 & 1 \\ 1 & -t & 1 \\ 1 & 1 & -t \end{vmatrix} = 0 \Rightarrow -t(t^2-1) - 1(-t-1) + 1(1+t) = 0 \Rightarrow (t+1)^2(t-2) = 0$ Thus $t = -1$ or $t = 2$.

3. Case analysis: (P) $|\vec{v}|^2 = 1$, which is item (2). (S) If $\alpha = \sqrt{2}$, then $\alpha^2 = 2$. Since $3\alpha^2=6$ in the case $t=2$ where $\alpha=\beta=\gamma$, we have $t=2$. Thus $t+3 = 5$, which is item (5). (Q) If $\alpha = \sqrt{3}$, we must be in the $t=-1$ case where $\alpha+\beta+\gamma=0$. Then $\beta+\gamma = -\sqrt{3}$ and $\beta^2+\gamma^2 = 6 - (\sqrt{3})^2 = 3$. Solving these gives $\gamma(\gamma+\sqrt{3})=0$, so $\gamma^2$ is $0$ or $3$. From options, item (1) matches. (R) If $\alpha = \sqrt{3}$, $(\beta+\gamma)^2 = (-\alpha)^2 = 3$, which is item (4).

Correct Option: A