The mean proportional of $\sqrt{3} + \sqrt{2}$ and $\sqrt{3} - \sqrt{2}$ is :

1. $\sqrt{5}$

2. 5

3. 1

4. 0

Let x be the mean proportion of $\sqrt{3} + \sqrt{2} \text{ and } \sqrt{3} - \sqrt{2}$ :

$\therefore \dfrac{\sqrt{3} + \sqrt{2}}{x} = \dfrac{x}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow (\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = x^2 \\[1em] \Rightarrow x^2 = (\sqrt{3})^2 - (\sqrt{2})^2 \quad [\because (a + b)(a - b) = a^2 - b^2] \\[1em] \Rightarrow x^2 = 3 - 2 \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = \sqrt{1} = \pm 1.$

Since, geometrical mean is always positive.

∴ x = 1.

Hence, Option 3 is the correct option.