If $$A = \{x \in R : |x-2| > 1\}$$, $$B = \{x \in R : \sqrt{x^2 - 3} > 1\}$$, $$C = \{x \in R : |x-4| \geq 2\}$$ and $$Z$$ is the set of all integers, then the number of subsets of the set $$(A \cap B \cap C)^c \cap Z$$ is _________.

We begin by translating each set description into interval form on the real number line.

For set $$A$$ we have the condition $$|x-2|>1$$. The definition of absolute value gives the equivalence

$$|x-2|>1 \;\Longrightarrow\; x-2<-1 \;\text{ or }\; x-2>1.$$

Simplifying each inequality,

$$x<1 \;\text{ or }\; x>3.$$

Hence

$$A=(-\infty,1)\cup(3,\infty).$$

For set $$B$$ the requirement is $$\sqrt{x^{2}-3}>1.$$ Because the square-root function is non-negative, we square both sides, knowing the direction of the inequality will not change:

$$\bigl(\sqrt{x^{2}-3}\bigr)^{2}>1^{2}\quad\Longrightarrow\quad x^{2}-3>1.$$

This rearranges to

$$x^{2}>4\quad\Longrightarrow\quad |x|>2.$$

Therefore

$$B=(-\infty,-2)\cup(2,\infty).$$

For set $$C$$ we are given $$|x-4|\ge 2$$. Using the same absolute-value definition,

$$|x-4|\ge 2 \;\Longrightarrow\; x-4\le-2 \;\text{ or }\; x-4\ge 2,$$

which simplifies to

$$x\le2 \;\text{ or }\; x\ge6.$$

Thus

$$C=(-\infty,2]\cup[6,\infty).$$

Now we find the intersection $$A\cap B$$. Writing the two interval unions together,

$$A=(-\infty,1)\cup(3,\infty),\qquad B=(-\infty,-2)\cup(2,\infty).$$

On the left half-line $$(-\infty,1)$$, the part common with $$B$$ is the portion further restricted by $$x<-2$$, giving $$(-\infty,-2).$$ On the right half-line $$(3,\infty)$$, every point already satisfies $$x>2$$, so the whole interval $$(3,\infty)$$ survives. Consequently

$$A\cap B=(-\infty,-2)\cup(3,\infty).$$

Next we intersect this result with $$C$$:

$$C=(-\infty,2]\cup[6,\infty).$$

The interval $$(-\infty,-2)$$ of $$A\cap B$$ obviously lies inside $$(-\infty,2]$$ of $$C$$, so it remains unchanged. The interval $$(3,\infty)$$ meets $$C$$ only where $$x\ge6$$, giving $$[6,\infty).$$ Therefore

$$A\cap B\cap C=(-\infty,-2)\cup[6,\infty).$$

We now form the complement of this set inside $$\mathbb R$$. Using the fact that the complement of a union is the union of complements, we remove both pieces from the real line:

$$\bigl(A\cap B\cap C\bigr)^{c}=\mathbb R\setminus\bigl((-\infty,-2)\cup[6,\infty)\bigr)=\,[ -2,6 ).$$

Here $$-2$$ is included because it was not in the intersection (the first interval was open at $$-2$$), while $$6$$ is excluded because it was contained in the intersection (the second interval was closed at $$6$$).

To finish, we intersect with $$Z$$, the set of all integers. The integers lying in $$[-2,6)$$ are

$$\{-2,-1,0,1,2,3,4,5\}.$$

We count their number:

$$n=8.$$

The question asks for the number of subsets of this finite set. A fundamental result of set theory states that a set with $$n$$ elements has $$2^{n}$$ distinct subsets (including the empty set and the set itself). Substituting $$n=8$$, we get

$$2^{8}=256.$$

So, the answer is $$256$$.