The problem asks to express the fraction $\frac{3\sqrt{2} - \sqrt{3}}{2\sqrt{3} - \sqrt{2}}$ in the form $\frac{\sqrt{m}}{\sqrt{n}}$, where $m$ and $n$ are whole numbers.

1. Problem Description

The problem asks to express the fraction $\frac{3\sqrt{2} - \sqrt{3}}{2\sqrt{3} - \sqrt{2}}$ in the form $\frac{\sqrt{m}}{\sqrt{n}}$, where $m$ and $n$ are whole numbers.

2. Solution Steps

First, we multiply the numerator and denominator by the conjugate of the denominator, which is $2\sqrt{3} + \sqrt{2}$:

$$\frac{3\sqrt{2} - \sqrt{3}}{2\sqrt{3} - \sqrt{2}} \cdot \frac{2\sqrt{3} + \sqrt{2}}{2\sqrt{3} + \sqrt{2}} = \frac{(3\sqrt{2} - \sqrt{3})(2\sqrt{3} + \sqrt{2})}{(2\sqrt{3} - \sqrt{2})(2\sqrt{3} + \sqrt{2})}$$

Now, we expand the numerator and denominator:

Numerator:

$(3\sqrt{2} - \sqrt{3})(2\sqrt{3} + \sqrt{2}) = 3\sqrt{2}(2\sqrt{3}) + 3\sqrt{2}(\sqrt{2}) - \sqrt{3}(2\sqrt{3}) - \sqrt{3}(\sqrt{2}) = 6\sqrt{6} + 3(2) - 2(3) - \sqrt{6} = 6\sqrt{6} + 6 - 6 - \sqrt{6} = 5\sqrt{6}$

Denominator:

$(2\sqrt{3} - \sqrt{2})(2\sqrt{3} + \sqrt{2}) = (2\sqrt{3})^2 - (\sqrt{2})^2 = 4(3) - 2 = 12 - 2 = 10$

So we have:

$$\frac{5\sqrt{6}}{10} = \frac{\sqrt{6}}{2}$$

Now, we want to express $\frac{\sqrt{6}}{2}$ in the form $\frac{\sqrt{m}}{\sqrt{n}}$.

Since $2 = \sqrt{4}$, we can write:

$$\frac{\sqrt{6}}{2} = \frac{\sqrt{6}}{\sqrt{4}} = \sqrt{\frac{6}{4}} = \sqrt{\frac{3}{2}}$$

To avoid a fraction inside the square root, we can manipulate the original expression further:

$$\frac{\sqrt{6}}{2} = \frac{\sqrt{6}}{\sqrt{4}} \cdot \frac{\sqrt{4}}{\sqrt{4}} = \frac{\sqrt{6\cdot4}}{4} = \frac{\sqrt{24}}{4} = \frac{\sqrt{24}}{\sqrt{16}}$$

This form satisfies the requirement of having whole numbers under the square root. However, looking at the given options, we see that we can directly write

$$\frac{5\sqrt{6}}{10} = \frac{\sqrt{25 \cdot 6}}{\sqrt{100}} = \frac{\sqrt{150}}{\sqrt{100}}$$

So, $m = 150$ and $n = 100$.

3. Final Answer

The final answer is (b) $\frac{\sqrt{150}}{\sqrt{100}}$.