The given problem involves rationalizing the denominator of the fraction $\frac{3}{\sqrt{2} + \sqrt{6}}$. Here's the process:

1. Rationalization: To rationalize the denominator $\sqrt{2} + \sqrt{6}$, we multiply both numerator and denominator by the conjugate of the denominator, $\sqrt{2} - \sqrt{6}$:

   $\frac{3}{\sqrt{2} + \sqrt{6}} \cdot \frac{\sqrt{2} - \sqrt{6}}{\sqrt{2} - \sqrt{6}}$

2. Simplify the Denominator: Using the difference of squares formula: $(\sqrt{2} + \sqrt{6})(\sqrt{2} - \sqrt{6}) = (\sqrt{2})^2 - (\sqrt{6})^2 = 2 - 6 = -4$

3. Simplify the Numerator: Multiply: $3(\sqrt{2} - \sqrt{6}) = 3\sqrt{2} - 3\sqrt{6}$

4. Final Fraction: Combine the simplified numerator and denominator: $\frac{3\sqrt{2} - 3\sqrt{6}}{-4} = \frac{-3\sqrt{2} + 3\sqrt{6}}{4}$

Thus, the rationalized form is: $\boxed{\frac{-3\sqrt{2} + 3\sqrt{6}}{4}}$

The correct answer is B.

Would you like further clarification or details? Here are some related questions to deepen understanding:

1. What is the purpose of multiplying by the conjugate to rationalize the denominator?
2. How does the difference of squares formula simplify radical expressions?
3. Can you provide another example involving a similar denominator?
4. How would this process change if the denominator were $\sqrt{2} - \sqrt{6}$?
5. Why is it often desirable to rationalize denominators in mathematics?

Tip: Always check the final denominator to ensure it no longer contains radicals after rationalization!