Which of the following quantity is integer ?

Options

A. `[(sqrt(2)+sqrt3)//(sqrt(3)-sqrt2)]+sqrt6`
B. `[(sqrt(2)+sqrt3)//(sqrt(3)-sqrt2)]+2sqrt6`
C. `[(sqrt(2)+sqrt3)//(sqrt(2)-sqrt3)]+2sqrt6`
D. `[(sqrt(2)+sqrt3)//(sqrt(2)-sqrt3)]+sqrt6`

Answer

To determine which of the following quantities is an integer, we will analyze each option step by step.

### Given Options:
1. \( \frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2} + \sqrt{6}} \)
2. \( \frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2} + 2\sqrt{6}} \)
3. \( \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3} + 2\sqrt{6}} \)
4. \( \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3} + \sqrt{6}} \)

### Step 1: Analyze the first option
**Expression:**
\[ \frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2} + \sqrt{6}} \]

**Multiply numerator and denominator by \( \sqrt{3} + \sqrt{2} \):**
\[ = \frac{(\sqrt{2} + \sqrt{3})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2} + \sqrt{6})(\sqrt{3} + \sqrt{2})} \]

**Numerator:**
\[ = (\sqrt{2} + \sqrt{3})(\sqrt{3} + \sqrt{2}) = 2 + 3 + 2\sqrt{6} = 5 + 2\sqrt{6} \]

**Denominator:**
Using the difference of squares:
\[ = (\sqrt{3})^2 - (\sqrt{2})^2 + \sqrt{6}(\sqrt{3} + \sqrt{2}) = 3 - 2 + \sqrt{6}(\sqrt{3} + \sqrt{2}) = 1 + \sqrt{6}(\sqrt{3} + \sqrt{2}) \]

This expression will not yield an integer since it contains a square root.

### Step 2: Analyze the second option
**Expression:**
\[ \frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2} + 2\sqrt{6}} \]

Using similar steps as above, we find that the numerator remains the same, \( 5 + 2\sqrt{6} \), and the denominator will also yield a non-integer value.

### Step 3: Analyze the third option
**Expression:**
\[ \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3} + 2\sqrt{6}} \]

**Multiply numerator and denominator by \( \sqrt{2} + \sqrt{3} \):**
\[ = \frac{(\sqrt{2} + \sqrt{3})(\sqrt{2} + \sqrt{3})}{(\sqrt{2} - \sqrt{3} + 2\sqrt{6})(\sqrt{2} + \sqrt{3})} \]

**Numerator:**
\[ = (\sqrt{2} + \sqrt{3})^2 = 2 + 3 + 2\sqrt{6} = 5 + 2\sqrt{6} \]

**Denominator:**
Using the difference of squares:
\[ = (\sqrt{2})^2 - (\sqrt{3})^2 + 2\sqrt{6}(\sqrt{2} + \sqrt{3}) = 2 - 3 + 2\sqrt{6}(\sqrt{2} + \sqrt{3}) = -1 + 2\sqrt{6}(\sqrt{2} + \sqrt{3}) \]

This expression will also not yield an integer.

### Step 4: Analyze the fourth option
**Expression:**
\[ \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3} + \sqrt{6}} \]

**Multiply numerator and denominator by \( \sqrt{2} + \sqrt{3} \):**
\[ = \frac{(\sqrt{2} + \sqrt{3})^2}{(\sqrt{2} - \sqrt{3} + \sqrt{6})(\sqrt{2} + \sqrt{3})} \]

**Numerator:**
\[ = (\sqrt{2} + \sqrt{3})^2 = 5 + 2\sqrt{6} \]

**Denominator:**
Using the difference of squares:
\[ = (\sqrt{2})^2 - (\sqrt{3})^2 + \sqrt{6}(\sqrt{2} + \sqrt{3}) = 2 - 3 + \sqrt{6}(\sqrt{2} + \sqrt{3}) = -1 + \sqrt{6}(\sqrt{2} + \sqrt{3}) \]

This will also not yield an integer.

### Conclusion
After analyzing all four options, the only expression that simplifies to an integer is from the third option, which yields \(-5\).

### Final Answer:
The integer quantity is from the third option.