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

For $A=C,A-C=\varphi$
$\Rightarrow \varphi \subseteq B$
But $A⊈B$
Hence, If $(A-C)\subseteq B,$ then $A\subseteq B$ is not true
Let $x\in C, \Rightarrow x\in \left(C\cup A\right)\cap \left(C\cup B\right)$
$\Rightarrow x\in \left(C\cup A\right)$ and $x\in \left(C\cup B\right)$
$\Rightarrow (x\in C$ or $x\in A)$ and $(x\in C$ or $x\in B)$
$\Rightarrow x\in C$ or $x\in \left(A\cap B\right)$
$\Rightarrow x\in C$ or $x\in C\left(asA\cup B\subseteq C\right)$
$\Rightarrow x\in C$
$\Rightarrow \left(C\cup A\right)\cap \left(C\cup B\right)\subseteq C$ .... (i)
Now $x\in C\Rightarrow x\in \left(C\cup A\right)$ and $x\in \left(C\cup B\right)$
$\Rightarrow x\in (C\cup A)\cap \left(C\cup B\right)$
$\Rightarrow C\subseteq (C\cup A)\cap \left(C\cup B\right)$ ..... (ii)
$\Rightarrow$ From (i) and (ii)
$C=\left(C\cup A\right)\cap \left(C\cup B\right)$
Hence, $(C\cup A)\cap (C\cup B)=C$ is true
Let $x\in A$ and $x\notin B$
$\Rightarrow x\in \left(A-B\right)$
$\Rightarrow x\in C$ $\left(asA-B\subseteq C\right)$
Let $x\in A$ and $x\in B$
$\Rightarrow x\in \left(A\cap B\right)$
$\Rightarrow x\in C$ $\left(asA\cap B\subseteq C\right)$
Hence $x\in A \Rightarrow x\in C$
$\Rightarrow A\subseteq C$
Hence, If $(A-B)\subseteq C,$ then $A\subseteq C$ is true
as $C\supseteq \left(A\cap B\right)$
$\Rightarrow B\cap C\supseteq \left(A\cap B\right)$
as $A\cap B\ne \varphi$
$\Rightarrow B\cap C\ne \varphi$
Hence, $B\cap C\ne \varphi$ is true

Alternate solution

Given 

$\varphi \ne A\cap B\subseteq C$

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) 

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

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

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

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

Option (c) $A-B=\left\{1,2,3\right\}\subseteq CA\subseteq C$ True

Option (d) $A-C=\varphi \subseteq BA\subseteq C$ Not correct

$\therefore$ Option (d) is correct