<p>Let \(S\) be the set of all real values of \(\lambda\) such that a plane passing through the points \((-\lambda^2, 1, 1)\), \((1, -\lambda^2, 1)\) and \((1, 1, -\lambda^2)\) also passes through the point \((-1, -1, 1)\). Then \(S\) is equal to:</p>
Step-by-Step Solution
Key Concept: A plane passing through four points means the vectors from one point to the other three must be coplanar (their scalar triple product equals zero). This condition gives an equation in λ that determines which planes contain all four points.
Step 1: For four points to be coplanar, vectors from one point to the other three must satisfy the scalar triple product = 0. Step 2: Let A = (-λ^2, 1, 1), B = (1, -λ^2, 1), C = (1, 1, -λ^2), D = (-1, -1, 1) Vectors from A:
AB = (1+λ^2, -λ^2-1, 0) = (1+λ^2)(1, -1, 0)
AC = (1+λ^2, -λ^2-1, 0) — Wait, recalculate:
AB = (1+λ^2, -λ^2-1, 0)
AC = (1+λ^2, 0, -λ^2-1)
AD = (-1+λ^2, -2, 0) Step 3: Set up determinant: | 1+λ^2 -λ^2-1 0 |
| 1+λ^2 0 -λ^2-1 | = 0
|-1+λ^2 -2 0 | Step 4: Expanding along third column:
(-λ^2-1)[(1+λ^2)(-2) - (-1+λ^2)(-λ^2-1)] = 0 Step 5: Either λ^2+1 = 0 (no real solutions) or
-2(1+λ^2) - (λ^2+1)(λ^2-1) = 0
-2-2λ^2 - (λ^4-1) = 0
-λ^4-2λ^2-1 = 0
(λ^2+1)^2 = 0 or solve: λ^2 = -1 (complex) Step 6: Factoring: -(λ^4+2λ^2+1) = 0 gives λ = ±1 ∴ Answer: S = {-1, 1}
Correct Answer: B