The number of real solutions of the equation $$3\left(x^2 + \frac{1}{x^2}\right) - 2\left(x + \frac{1}{x}\right) + 5 = 0$$, is

Let $$t = x + \frac{1}{x}$$. Then $$x^2 + \frac{1}{x^2} = t^2 - 2$$.

The equation becomes:

$$3(t^2 - 2) - 2t + 5 = 0$$

$$3t^2 - 6 - 2t + 5 = 0$$

$$3t^2 - 2t - 1 = 0$$

$$(3t + 1)(t - 1) = 0$$

$$t = -\frac{1}{3}$$ or $$t = 1$$

Case 1: $$x + \frac{1}{x} = 1$$

$$x^2 - x + 1 = 0$$

Discriminant = $$1 - 4 = -3 < 0$$. No real solutions.

Case 2: $$x + \frac{1}{x} = -\frac{1}{3}$$

$$3x^2 + x + 3 = 0$$

Discriminant = $$1 - 36 = -35 < 0$$. No real solutions.

Therefore, the total number of real solutions is 0.

The correct answer is Option 2: 0.