Let $$f(x) = 3(x^2 - 2)^3 + 4$$, $$x \in \mathbb{R}$$. Then which of the following statements are true?
P: $$x = 0$$ is a point of local minima of f
Q: $$x = \sqrt{2}$$ is a point of inflection of f
R: $$f'$$ is increasing for $$x > \sqrt{2}$$

We have $$f(x) = 3(x^2 - 2)^3 + 4$$. We compute $$f'(x) = 3 \cdot 3(x^2 - 2)^2 \cdot 2x = 18x(x^2 - 2)^2$$.

Setting $$f'(x) = 0$$: either $$x = 0$$ or $$x^2 - 2 = 0$$, giving $$x = 0, \pm\sqrt{2}$$.

Statement P: $$x = 0$$ is a point of local minima. We observe that $$(x^2 - 2)^2 \geq 0$$ always, so the sign of $$f'(x)$$ is determined by the sign of $$x$$. For $$x$$ slightly less than 0, $$f'(x) < 0$$ (decreasing), and for $$x$$ slightly greater than 0, $$f'(x) > 0$$ (increasing). Since $$f'$$ changes sign from negative to positive, $$x = 0$$ is indeed a local minimum. P is true.

Statement Q: $$x = \sqrt{2}$$ is a point of inflection. We compute $$f''(x)$$. Differentiating $$f'(x) = 18x(x^2 - 2)^2$$: $$f''(x) = 18(x^2 - 2)^2 + 18x \cdot 2(x^2 - 2) \cdot 2x = 18(x^2 - 2)\big[(x^2 - 2) + 4x^2\big] = 18(x^2 - 2)(5x^2 - 2)$$.

At $$x = \sqrt{2}$$: $$f''(\sqrt{2}) = 18(2 - 2)(10 - 2) = 0$$. We check for a sign change: for $$x$$ slightly less than $$\sqrt{2}$$, $$x^2 - 2 < 0$$ while $$5x^2 - 2 > 0$$, so $$f'' < 0$$. For $$x$$ slightly greater than $$\sqrt{2}$$, both factors are positive, so $$f'' > 0$$. Since $$f''$$ changes sign, $$x = \sqrt{2}$$ is a point of inflection. Q is true.

Statement R: $$f'$$ is increasing for $$x > \sqrt{2}$$. This requires $$f''(x) > 0$$ for $$x > \sqrt{2}$$. We have $$f''(x) = 18(x^2 - 2)(5x^2 - 2)$$. For $$x > \sqrt{2}$$: $$x^2 > 2$$ so $$x^2 - 2 > 0$$, and $$5x^2 > 10 > 2$$ so $$5x^2 - 2 > 0$$. Hence $$f''(x) > 0$$, confirming $$f'$$ is increasing. R is true.

Since all three statements P, Q, and R are true, the correct answer is Option D.

Hence, the correct answer is Option D.