The set of all values of λ for which the system of linear equations x + y + z = λx, x + 2y + z = λy, -x - y = λz has a non-trivial solution.

x + y + z = λx
(1 - λ)x + y + z = 0 (1)
x + 2y + z = λy
x + (2 - λ)y + z = 0
-x - y = λz
x + y + λz = 0
Since the system of given linear equations has a non-trivial solution.
Therefore,
| (1 - λ) 1 1 |
| 1 (2 - λ) 1 | = 0
| 1 1 λ |
((1 - λ)(λ(2 - λ) - 1) - (λ - 1) + (1 - (2 - λ))) = 0
(1 - λ)(2λ - λ² - 1) - (λ - 1) + (λ - 1) = 0
(1 - λ)(2λ - λ² - 1) = 0
(1 - λ)(-(λ² - 2λ + 1)) = 0
(1 - λ)(- (λ - 1)²) = 0
(1 - λ)(λ - 1)² = 0
(1 - λ)³ = 0
λ = 1
Thus, the set of values of λ contains exactly one element.
Hence the correct answer is (B).