Let A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}.

Solution:

Given: A = {a, b, c, d,}, B = {c, d, e} and C = {d, e, f, g}

(i) Need to prove: $A \times(B \cap C)=(A \times B) \cap(A \times C)$

Left hand side

$(B \cap C)=\{d, e\}$

$\Rightarrow A \times(B \cap C)=\{(a, d),(a, e),(b, d),(b, e),(c, d),(c, e),(d, d),(d, e)\}$

Right hand side

$(A \times B)=\{(a, c),(a, d),(a, e),(b, c),(b, d),(b, e),(c, c),(c, d),(c, e),(d, c),(d, d),(d, e)\}$

$(A \times C)=\{(a, d),(a, e),(a, f),(a, g),(b, d),(b, e),(b, f),(b, g),(c, d),(c, e),(c, f),(c, g)$

$(\mathrm{d}, \mathrm{d}),(\mathrm{d}, \mathrm{e}),(\mathrm{d}, \mathrm{f}),(\mathrm{d}, \mathrm{g})\}$

Now,

$(A \times B) \cap(A \times C)=\{(a, d),(a, e),(b, d),(b, e),(c, d),(c, e),(d, d),(d, e)\}$

Here, right hand side and left hand side are equal.

That means, $A \times(B \cap C)=(A \times B) \cap(A \times C)$ [Proved]

(ii) Need to prove: $A \times(B-C)=(A \times B)-(A \times C)$

Left hand side,

$(B-C)=\{c\}$

$\Rightarrow A \times(B-C)=\{(a, c),(b, c),(c, c),(d, c)\}$

Right hand side,

$(A \times B)=\{(a, c),(a, d),(a, e),(b, c),(b, d),(b, e),(c, c),(c, d),(c, e),(d, c),(d, d),(d, e)\}$

$(A \times C)=\{(a, d),(a, e),(a, f),(a, g),(b, d),(b, e),(b, f),(b, g),(c, d),(c, e),(c, f),(c, g)$

$(d, d),(d, e),(d, f),(d, g)\}$

Therefore, $(A \times B)-(A \times C)=\{(a, c),(b, c),(c, c),(d, c)\}$

Here, right hand side and left hand side are equal.

That means, $A \times(B-C)=(A \times B)-(A \times C)$ [Proved]

(iii) Need to prove:

$(A \times B) \cap(B \times A)=(A \cap B) \times(A \cap B)$

Left hand side,

$(A \times B)=\{(a, c),(a, d),(a, e),(b, c),(b, d),(b, e),(c, c),(c, d),(c, e),(d, c),(d, d),(d, e)\}$

$(B \times A)=\{(c, a),(c, b),(c, c),(c, d),(d, a),(d, b),(d, c),(d, d),(e, a),(e, b),(e, c),(e, d)\}$

Now, $(A \times B) \cap(B \times A)=\{(c, c),(c, d),(d, c),(d, d)\}$

Right hand side,

$(A \cap B)=\{c, d\}$

So, $(A \cap B) \times(A \cap B)=\{(c, c),(c, d),(d, c),(d, d)\}$

Here, right hand side and left hand side are equal

That means, $(A \times B) \cap(B \times A)=(A \cap B) \times(A \cap B)[$ Proved $]$