A = {1, 2, 3}
R₁ = {(1,2), (1,3), (1, 1), (2, 2), (3, 3), (2, 1), (3, 1)} is the only relation on {1, 2, 3} which is reflexive, symmetric but not transitive and is such that (1, 2), (1, 3) ∈ R₁.
∴ (A) is correct answer.

A is the set of all books in a library of a college.
R = {(x,y) : x and y have same number of pages}
Since (x, x) ∈ R as x and x have the same number of pages ∀ x ∈ A.
∴ R is reflexive.
Also (x, y) ∈ R
⇒ x and y have the same number of pages ⇒ y and x have the same number of pages
⇒ (y, x) ∈ R
∴ R is symmetric.
Now, (x, y) ∈ R and (y, z) ∈ R.
⇒ x and y have the same number of pages and y and z have the same number of pages
⇒ x and z have he same number of pages ⇒ (x, z) ∈ R ∶ R is transitive.

R = {(T₁, T₂) : T₁ is congruent to T₂}
(i) Since every triangle is congruent to itself
∴ R is reflexive.
(ii) Also (T₁, T₂) ∈ R ⇒ T₁ is congruent to T₂ ⇒ T₂ is congruent to T₁ (T₂,T₁) ∈ R
(T₁,T₂) ∈ R ⇒ (T₂,T₁) ∈ R ⇒ R is symmetric.
(iii) Again (T₁, T₂), (T₂, T₃) ∈ R ⇒ T₁ i is congruent to T₂ and T₂ is congruent to T₃ ∴ T₁ is congruent to T₃ ∴ (T₁,T₃) ∈ R (T₁,T₂), (T₂,T₃) ∈ R ⇒ (T₁,T₃) ∈ R ∴ R is transitive.
From (i), (ii), (iii), it is clear that R is reflexive, symmetric and transitive
∴ R is an equivalence relation.

R = {(a, b) : a = b – 2, b > 6}
∴ (a, b) ∈ R ⇒ a = b – 2 where b > 6 ∴ (6, 8) ∈ R as 6 = 8 – 2 where b = 8 > 6 ∴ (C) is correct answer.

R = {(T₁, T₂) : T₁ is similar to T₂}
Since every triangle is similar to itself
∴ R is reflexive.
Also (T₁ T₂) ∈ R ⇒ T₁ is similar to T₂ ⇒ T₂ is similar to T₁ ∴ (T₂,T₁) ⇒ R
∴ (T₁,T₂) ∈ R ⇒ (T₂,T₁) ∈ R ⇒ R is symmetric.
Again (T₁, T₂), (T₂, T₃) ∈ R
⇒ T₁ is similar to T₂ and T₂ is similar to T₃
∴ T₁ is similar to T₃ ⇒ (T₁,T₃) ∈ R ∴ (T₁, T₂), (T₂,T₃) ∈ R ⇒ (T₁, T₃) ∈ R ∴ R is transitive.
∴ R is reflexive, symmetric and transitive ∴ R is an equivalence relation.
Now T₁, T₂, T₃ are triangles with sides 3, 4, 5 ; 5, 12, 13 and 6, 8, 10.

Since         

∴ T₁ is similar to T₃ i.e. T₃ is similar to T₁.
No two other triangles are similar.