The number of integer solutions $x$ of the equation below

$(12x -1)(6x - 1)(4x -1)(3x - 1) = 330$ is

(A): $0$, (B): $1$, (C): $2$, (D): $3$, (E): None of the above.
1. Start with the given equation:
   \[
   (12x - 1)(6x - 1)(4x - 1)(3x - 1) = 330
   \]

2. Notice that the equation can be rearranged in any order due to the commutative property of multiplication:
   \[
   (12x - 1)(3x - 1)(6x - 1)(4x - 1) = 330
   \]

3. To simplify the problem, let's introduce a substitution. Let:
   \[
   y = x(12x - 5)
   \]

4. Then, the equation becomes:
   \[
   (3y + 1)(2y + 1) = 330
   \]

5. Factorize 330:
   \[
   330 = 2 \cdot 3 \cdot 5 \cdot 11
   \]

6. Since \((3y + 1)\) and \((2y + 1)\) are factors of 330, we need to find pairs of factors that satisfy the equation. We test different pairs of factors:
   - \( (3y + 1) = 2 \cdot 11 = 22 \) and \( (2y + 1) = 3 \cdot 5 = 15 \)
   - \( (3y + 1) = 3 \cdot 5 = 15 \) and \( (2y + 1) = 2 \cdot 11 = 22 \)

7. Solve for \(y\) in each case:
   - For \( (3y + 1) = 22 \):
     \[
     3y + 1 = 22 \implies 3y = 21 \implies y = 7
     \]
   - For \( (2y + 1) = 15 \):
     \[
     2y + 1 = 15 \implies 2y = 14 \implies y = 7
     \]

8. Since both cases give \( y = 7 \), we substitute back to find \(x\):
   \[
   y = x(12x - 5) = 7
   \]

9. Solve the quadratic equation:
   \[
   12x^2 - 5x - 7 = 0
   \]

10. Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
    \[
    a = 12, \quad b = -5, \quad c = -7
    \]
    \[
    x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 12 \cdot (-7)}}{2 \cdot 12}
    \]
    \[
    x = \frac{5 \pm \sqrt{25 + 336}}{24}
    \]
    \[
    x = \frac{5 \pm \sqrt{361}}{24}
    \]
    \[
    x = \frac{5 \pm 19}{24}
    \]

11. This gives two potential solutions:
    \[
    x = \frac{24}{24} = 1 \quad \text{and} \quad x = \frac{-14}{24} = -\frac{7}{12}
    \]

12. Since we are looking for integer solutions, only \( x = 1 \) is valid.

The final answer is \( \boxed{1} \)