There are an infinite number of 3 element real number sets with any given real mean and any given positive real standard diviation.

Without loss of generality lets assume that $\mu = 0$ and $\sigma = 1$. Once we have a soloution set for these parameters we can find one for any parameters by scaling it to get the desired standard deviation, then shifting it to get the desired mean

$$x + y + z = 0$$

$$\frac{x^2 + y^2 + z^2}{3} - 0 = 1$$

$$x^2 + y^2 + z^2 = 3$$

We have two equations and 3 unknowns, so lets treat x as a parameter and solve for y and z.

$$y = -x -z$$

$$x^2 + (-x -z)^2 + z^2 = 3$$

$$x^2 + (x^2 + 2xz +z^2) + z^2 = 3$$

$$2z^2 + 2xz + (2x^2 -3) = 0$$

$$z = \frac{-2x\pm\sqrt{(2x)^2-8(2x^2-3)}}{4}$$

$$z = \frac{-2x\pm\sqrt{4x^2-16x^2+24}}{4}$$

$$z = \frac{-2x\pm\sqrt{-12x^2+24}}{4}$$

For this to have real soloutions we need the following inequality to hold.

$$-12x^2+24 \geqslant 0$$

$$12x^2 \leqslant 24 $$

$$x^2 \leqslant 2 $$

$$- \sqrt{2} \leqslant x \leqslant \sqrt{2} $$

There are an infinite number of $x$ values that satisfy this inequality, therefore there are an infinite number of sets of 3 real numbers with $\mu = 0$ and $\sigma = 1$ therefore there are an infinite number of sets of three real numbers with any given real mean and any given positive* real standard deviation.

* Negative standard diviations don't make any sense, and a zero standard deviation means all numbers are equal.