If A = x ∈ R: | x - 2 | > 1 , B = \ x ∈ R: x ^ 2 - 3 > 1\ , C = x ∈...

For First set A = { x \in R: | x - 2 | > 1 }$</p><p>$⇒$Value which A$ contains lies from $x \in \left(- \infty , 1\right) \cup \left(3, \infty\right)</p><p>For second set$B = \left\{x \in R: \sqrt{{x}^{2} - 3} > 1\right\}$</p><p>$⇒$Value which$B$contains lies from$x \in \left(- \infty , - 2\right) \cup \left(2, \infty\right)

For third set $C = { x \in R: | x - 4 | ⩾ 2 }$

$⇒$ Value which $C$ contains lies from $x \in ( - \infty , 2 ] \cup [ 6, \infty )$

Now, \left(- \infty , - 2\right) \cup [ 6, \infty ) \in \left(A \cap B \cap C\right)

Therefore, ${\left(A \cap B \cap C\right)}^{c} \cap Z = \left\{- 2, - 1, 0, 1, 2, 3, 4, 5\right\}</p><p>Number$ of Subsets of above set= {2}^{8} = 256$</p></blockquote>$