i. {(x, y): y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}

When x = 1, we have y = 3(1) = 3

When x = 2, we have y = 3(2) = 6

When x = 3, we have y = 3(3) = 9

Thus, R = {(1, 3), (2, 6), (3, 9)}

Every element of set x has an ordered pair in the relation and no two ordered pairs have the same first component but different second components.

Hence, the given relation R is a function.

ii. {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6}

When x = 1, we have y > 1 + 1 or y > 2 ⇒ y = {4, 6}

When x = 2, we have y > 2 + 1 or y > 3 ⇒ y = {4, 6}

Thus, R = {(1, 4), (1, 6), (2, 4), (2, 6)}

Every element of set x has an ordered pair in the relation. However, two ordered pairs (1, 4) and (1, 6) have the same first component but different second components.

Hence, the given relation R is not a function.

iii. {(x, y): x + y = 3, x, y ∈ {0, 1, 2, 3}}

When x = 0, we have 0 + y = 3 ⇒ y = 3

When x = 1, we have 1 + y = 3 ⇒ y = 2

When x = 2, we have 2 + y = 3 ⇒ y = 1

When x = 3, we have 3 + y = 3 ⇒ y = 0

Thus, R = {(0, 3), (1, 2), (2, 1), (3, 0)}

Every element of set x has an ordered pair in the relation and no two ordered pairs have the same first component but different second components.

Hence, the given relation R is a function.