Let A, B and C be sets such that ϕ ≠ A B ⊆ C . Then which of the...

For $A = C, A - C = ϕ$
$⇒ ϕ ⊆ B$
But $A ⊈ B$
Hence, If $( A - C ) ⊆ B,$ then $A ⊆ B$ is not true
Let x \in C, ⇒ x \in \left(C \cup A\right) \cap \left(C \cup B\right)$<br />$⇒ x \in \left(C \cup A\right)$and$x \in \left(C \cup B\right)$<br />$⇒ ( x \in C$or$x \in A )$and$( x \in C$or$x \in B )$<br />$⇒ x \in C$or$x \in \left(A \cap B\right)
$⇒ x \in C$ or $x \in C \left(asA \cup B ⊆ C\right)$
$⇒ x \in C$
$⇒ \left(C \cup A\right) \cap \left(C \cup B\right) ⊆ C$ .... (i)
Now $x \in C ⇒ x \in \left(C \cup A\right)$ and x \in \left(C \cup B\right)$<br />$⇒ x \in ( C \cup A ) \cap \left(C \cup B\right)$<br />$⇒ C ⊆ ( C \cup A ) \cap \left(C \cup B\right) ..... (ii)
$⇒$ From (i) and (ii)
C = \left(C \cup A\right) \cap \left(C \cup B\right)$<br />Hence,$( C \cup A ) \cap ( C \cup B ) = C$is true<br />Let$x \in A$and$x \notin B$<br />$⇒ x \in \left(A - B\right)
$⇒ x \in C \left(asA - B ⊆ C\right)$
Let $x \in A$ and $x \in B$
$⇒ x \in \left(A \cap B\right)$
⇒ x \in C \left(asA \cap B ⊆ C\right)$<br />Hence$x \in A ⇒ x \in C$<br />$⇒ A ⊆ C$<br />Hence, If$( A - B ) ⊆ C,$then$A ⊆ C$is true<br />as$C ⊇ \left(A \cap B\right)$<br />$⇒ B \cap C ⊇ \left(A \cap B\right)
as $A \cap B \ne ϕ$
$⇒ B \cap C \ne ϕ$
Hence, $B \cap C \ne ϕ$ is true

Alternate solution

Given~

$ϕ \ne A \cap B ⊆ C$

Let answer that

$C = \left\{1, 2, 3, 4, 5, 6\right\}$

$A = \left\{1, 2, 3\right\}$

$B = \left\{3, 4, 5\right\}$

$A \cap B = \left\{3\right\}$

Option (a)~

$B \cap C = \left\{3, 4, 5\right\} \ne ϕ$

Option (b)~$C \cup A = \left\{1, 2, 3, 4, 5, 6\right\}$

$C \cup B = \left\{1, 2, 3, 4, 5, 6\right\}$

\left(C \cup A\right) \cap \left(C \cup B\right) = \left\{1, 2, 3, 4, 5, 6\right\} = C$</p><p>Option (c)~$A - B = \left\{1, 2, 3\right\} ⊆ CA ⊆ C$~True</p><p>Option (d)~$A - C = ϕ ⊆ BA ⊆ C$~Not correct</p><p>$∴& Option (d) is correct&