To solve the problem, we need to determine the sets \( A \), \( B \), and \( C \) based on the given conditions, and then find the number of integral values in the intersection of these sets and their complement.
### Step 1: Define Set \( A \)
Set \( A \) is defined as:
\[
A = \{ x : |x - 2| > 1 \}
\]
This can be split into two cases:
1. **Case 1**: \( x - 2 > 1 \)
\[
x > 3 \quad \Rightarrow \quad x \in (3, \infty)
\]
2. **Case 2**: \( -(x - 2) > 1 \)
\[
-x + 2 > 1 \quad \Rightarrow \quad x < 1 \quad \Rightarrow \quad x \in (-\infty, 1)
\]
Combining both cases, we have:
\[
A = (-\infty, 1) \cup (3, \infty)
\]
### Step 2: Define Set \( B \)
Set \( B \) is defined as:
\[
B = \{ x : \sqrt{x^2 - 3} > 1 \}
\]
Squaring both sides, we get:
\[
x^2 - 3 > 1 \quad \Rightarrow \quad x^2 > 4 \quad \Rightarrow \quad x > 2 \text{ or } x < -2
\]
Thus, we can write:
\[
B = (-\infty, -2) \cup (2, \infty)
\]
### Step 3: Define Set \( C \)
Set \( C \) is defined as:
\[
C = \{ x : |x - 4| \geq 2 \}
\]
This can also be split into two cases:
1. **Case 1**: \( x - 4 \geq 2 \)
\[
x \geq 6 \quad \Rightarrow \quad x \in [6, \infty)
\]
2. **Case 2**: \( -(x - 4) \geq 2 \)
\[
-x + 4 \geq 2 \quad \Rightarrow \quad x \leq 2 \quad \Rightarrow \quad x \in (-\infty, 2]
\]
Combining both cases, we have:
\[
C = (-\infty, 2] \cup [6, \infty)
\]
### Step 4: Find \( A \cap B \cap C \)
Now we need to find the intersection \( A \cap B \cap C \).
1. **Intersection of \( A \) and \( B \)**:
\[
A \cap B = \left((- \infty, 1) \cup (3, \infty)\right) \cap \left((- \infty, -2) \cup (2, \infty)\right)
\]
This gives:
\[
A \cap B = (-\infty, -2) \cup (3, \infty)
\]
2. **Intersection with \( C \)**:
\[
C = (-\infty, 2] \cup [6, \infty)
\]
Now we find:
\[
(A \cap B) \cap C = \left((- \infty, -2) \cup (3, \infty)\right) \cap \left((- \infty, 2] \cup [6, \infty)\right)
\]
This results in:
\[
(A \cap B) \cap C = (-\infty, -2) \cup [6, \infty)
\]
### Step 5: Find the Complement
Now we need the complement of \( A \cap B \cap C \):
\[
(A \cap B \cap C)' = \mathbb{R} \setminus \left((- \infty, -2) \cup [6, \infty)\right)
\]
This gives:
\[
(A \cap B \cap C)' = (-2, 6)
\]
### Step 6: Find the Intersection with \( Z \)
Now we find the intersection with the set of integers \( Z \):
\[
(A \cap B \cap C)' \cap Z = (-2, 6) \cap Z
\]
The integers in this interval are:
\[
-1, 0, 1, 2, 3, 4, 5
\]
Counting these gives us 7 integers.
### Final Answer
Thus, the value of \( k \) is:
\[
\boxed{7}
\]
