To solve the inequality $|3 - 2x| > 11$, we need to break it down into two separate cases, since the absolute value inequality can be expressed as:

$3 - 2x > 11 \quad \text{or} \quad 3 - 2x < -11.$

Let's solve each case individually:

Case 1: $3 - 2x > 11$

1. Subtract 3 from both sides: $-2x > 8$
2. Divide by -2 (remember to flip the inequality sign when dividing by a negative number): $x < -4$

Case 2: $3 - 2x < -11$

1. Subtract 3 from both sides: $-2x < -14$
2. Divide by -2 (again, flip the inequality sign): $x > 7$

Solution Set

The combined solution is $x < -4$ or $x > 7$. In interval notation, this is:

$(-\infty, -4) \cup (7, \infty)$

Thus, the correct answer is option A, with the solution set being $(-\infty, -4) \cup (7, \infty)$.

Would you like more details or have any questions?

Here are some related questions to explore:

1. How do you solve absolute value inequalities in general?
2. How would the solution change if the inequality were $|3 - 2x| \leq 11$?
3. What happens if the inequality were $|3 - 2x| < -11$? Is there any real solution?
4. How would you graph the solution set on a number line?
5. Can you solve $|2x - 3| > 5$ using a similar method?

Tip: When solving absolute value inequalities, remember to split them into two cases based on the definition of absolute value, and always flip the inequality when multiplying or dividing by a negative number.