Let S be the set of all real values of λ such that a plane passing through the points (−λ²,1,1), (1,−λ²,1) and (1,1,−λ²) also passes through the point (-1, -1, 1). Then S is equal to:

Let A = (−λ², 1, 1), B = (1, −λ², 1), C = (1, 1, −λ²), and D = (−1, −1, 1).
The points A, B, C, and D are coplanar if and only if the scalar triple product of the vectors AB, AC, and AD is zero.
AB = B - A = (1 + λ², −λ² − 1, 0)
AC = C - A = (1 + λ², 0, −λ² − 1)
AD = D - A = (1 − λ², −2, 0)
The scalar triple product is given by the determinant:
| 1+λ² −λ²−1 0 |
| 1+λ² 0 −λ²−1 |
| 1−λ² −2 0 |
Expanding the determinant, we get:
(1 + λ²)[0 − (−2)(−λ² − 1)] − (−λ² − 1)[0 − (1 − λ²)(−λ² − 1)] + 0 = 0
(1 + λ²)[−2(λ² + 1)] + (λ² + 1)[(1 − λ²)(λ² + 1)] = 0
(1 + λ²)[−2(λ² + 1) + (1 − λ²)(λ² + 1)] = 0
(1 + λ²)(λ² + 1)[−2 + 1 − λ²] = 0
(1 + λ²)(λ² + 1)(−1 − λ²) = 0
(1 + λ²)(λ² + 1)(λ² + 1) = 0
(1 + λ²)(λ² + 1)² = 0
This implies that 1 + λ² = 0, which gives λ² = −1, which has no real solutions.
If λ² + 1 = 0, then λ² = −1. This has no real solutions.
However, there must be a mistake in the calculation. Let's recalculate the scalar triple product using the determinant:
| 1+λ² −1−λ² 0 |
| 1+λ² 0 −1−λ² |
| 1−λ² −2 0 |
Expanding along the third column:
(1+λ²)(2(1+λ²))+(1+λ²)(1−λ²)(1+λ²) = 0
(1+λ²)[2(1+λ²) + (1−λ²)(1+λ²)] = 0
(1+λ²)(1+λ²)(2 + 1 − λ²) = 0
(1+λ²)²(3 − λ²) = 0
Therefore, either 1 + λ² = 0 (no real solution) or 3 − λ² = 0, which means λ² = 3.
Thus, λ = ±√3.
Therefore, S = {√3, −√3}