Let α, β and γ be real numbers. Consider the following system of linear equations
x + 2y + z = 7
x + αz = 11
2x – 3y + βz = γ
Match each entry in List-I to the correct entries in List-II.

List-I List-II
(P) If \(\beta=\frac{1}{2}(7 \alpha-3)\) and γ = 28, then the system has (1) a unique solution
(Q) If \(\beta=\frac{1}{2}(7 \alpha-3)\) and γ ≠ 28, then the system has (2) no solution
(R) If \(\beta\ne\frac{1}{2}(7 \alpha-3)\) where α = 1 and γ ≠ 28, then the
system has
(3) infinitely many solutions
(S) If \(\beta\ne\frac{1}{2}(7 \alpha-3)\) where α = 1 and γ = 28, then the
system has
(4) x = 11, y = –2 and z = 0 as a solution
  (5) x = –15, y = 4 and z = 0 as a solution

The correct option is:

1
(P) → (3), (Q) → (2), (R) → (1), (S) → (4)
2
(P) → (3), (Q) → (2), (R) → (5), (S) → (4)
3
(P) → (2), (Q) → (1), (R) → (4), (S) → (5)
4
(P) → (2), (Q) → (1), (R) → (1), (S) → (3)

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