Question 16

**Analysis:** From u×v = w and v×w = u, u ⊥ v, u ⊥ w, v ⊥ w, so |w| = |u||v| and |u| = |v||w| ⇒ |v|² = 1 and |u|² = |w|² = 6. The homogeneous system has nontrivial solution when det = −(t+1)²(t−2) = 0 ⇒ t = −1 or t = 2. - t = 2: α = β = γ, 3α² = 6 ⇒ α = √2, so t + 3 = 5 → (S)→(5). - t = −1: α+β+γ = 0 with u ⊥ w (α+β−2γ = 0) ⇒ γ = 0, β = −α, 2α² = 6 ⇒ α = √3. Then γ² = 0 → (Q)→(1), (β+γ)² = 3 → (R)→(4). (P)→(2).

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