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

1. Problem Description

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

2. Solution Steps

We have to rationalize the denominator of the given expression: $\frac{3\sqrt{2} - \sqrt{3}}{2\sqrt{3} - \sqrt{2}}$.

To rationalize the denominator, we multiply both the numerator and the 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})}

$$$$

Expanding the numerator:

$$$$

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

$$$$

Expanding the denominator (using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$):

$$$$

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

$$$$

So the expression becomes:

$$$$

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

$$$$

Now we need to express this in the form $\frac{\sqrt{m}}{\sqrt{n}}$.

Since $2 = \sqrt{4}$, we can rewrite the expression as:

$$$$

\frac{\sqrt{6}}{\sqrt{4}}

$$$$

We need to get the options provided in the form $\frac{\sqrt{m}}{\sqrt{n}}$ and check if they equal to $\frac{\sqrt{6}}{2}$ or $\frac{\sqrt{6}}{\sqrt{4}}$.

(a) $\frac{\sqrt{6}}{\sqrt{10}}$

(b) $\frac{\sqrt{150}}{\sqrt{100}} = \frac{\sqrt{25 \cdot 6}}{\sqrt{100}} = \frac{5\sqrt{6}}{10} = \frac{\sqrt{6}}{2}$

(c) $\frac{2\sqrt{6}}{\sqrt{10}}$

(d) $\frac{5\sqrt{6}}{\sqrt{10}}$

The correct option is (b) because $\frac{\sqrt{150}}{\sqrt{100}} = \frac{\sqrt{150}}{10} = \frac{5\sqrt{6}}{10} = \frac{\sqrt{6}}{2} = \frac{\sqrt{6}}{\sqrt{4}}$.

3. Final Answer

(b) $\frac{\sqrt{150}}{\sqrt{100}}$