The set \((\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \cap\left(\mathrm{A} \cap \mathrm{B}^{\prime} \cap \mathrm{C}^{\prime}\right)^{\prime} \cap \mathrm{C}^{\prime}\) equals (a) \(\mathrm{B} \cap \mathrm{C}^{\prime}\) (b) \(\mathrm{B} \cup \mathrm{C}^{\prime}\) (c) \(\mathrm{A} \cap \mathrm{C}\) (d) \(\mathrm{A} \cup \mathrm{C}\)

To simplify the expression, apply De Morgan's law to \( (\mathrm{A}\cap \mathrm{B}' \cap \mathrm{C}')' \). This law states that: \( (A \cap B)’ = A’ \cup B’ \). Therefore, we have: \((\mathrm{A}\cap \mathrm{B}' \cap \mathrm{C}')' = (\mathrm{A}')\cup(\mathrm{B}'')\cup(\mathrm{C}'') \) As we know \( P''=P \), the expression becomes: \((\mathrm{A}\cap \mathrm{B}' \cap \mathrm{C}')' = (\mathrm{A}')\cup \mathrm{B} \cup \mathrm{C} \)

Now, utilize the distributive property of intersections to simplify further. The distributive property states that: \((A ∩ (B ∪ C)) = (A ∩ B) ∪ (A ∩ C)\). Thus, the given expression is: \((\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}'\) To distribute the intersection: \((\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}' = (\mathrm{A}\cap (\mathrm{A}'\cup \mathrm{B}\cup \mathrm{C})\cap \mathrm{C}' )\cup (\mathrm{B}\cap (\mathrm{A}'\cup \mathrm{B}\cup \mathrm{C})\cap \mathrm{C}' )\cup (\mathrm{C}\cap (\mathrm{A}'\cup \mathrm{B}\cup \mathrm{C})\cap \mathrm{C}' )\).

We will now use several set properties to simplify the expression. The properties are: 1. \(A \cap A’ = ∅\) 2. \(A \cap ∅ = ∅\) 3. \(A \cap B = B \cap A\) 4. \(A \cap (B ∪ C) = (A ∩ B) ∪ (A ∩ C)\) Apply these properties to the expression: \((\mathrm{A} \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}') = ∅ \) \((\mathrm{B} \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}') = (\mathrm{B} \cap \mathrm{A}') \cup (\mathrm{B}\cap\mathrm{B})\cup (\mathrm{B} \cap \mathrm{C})\cap \mathrm{C'\) ) Following set properties 1 and 2: \((\mathrm{B} \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}') = (\mathrm{B} \cap \mathrm{A}') \cup ∅ \cup (\mathrm{B} \cap \mathrm{C}\cap \mathrm{C'})\) This simplifies to: \((\mathrm{B} \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}') = (\mathrm{B} \cap \mathrm{A}') \) \((\mathrm{C}\cap (\mathrm{A}'\cup \mathrm{B}\cup \mathrm{C})\cap \mathrm{C}' ) = ∅\) Thus, we have: \((\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}' = (\mathrm{B} \cap \mathrm{A}')\)

Comparing the simplified expression with the given options, we see that: \((\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \cap (\mathrm{A}' \cup \mathrm{B} \cup \mathrm{C})\cap \mathrm{C}' = (\mathrm{B} \cap \mathrm{A}')\) Which is equal to option (a): \((\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \cap\left(\mathrm{A}\cap \mathrm{B}' \cap \mathrm{C}'\right)' \cap \mathrm{C}' = \mathrm{B} \cap \mathrm{C}'\)