The sum of the squares of the roots of $$|x - 2|^2 + |x - 2| - 2 = 0$$ and the squares of the roots of $$x^2 - 2|x - 3| - 5 = 0$$, is

We need to find the sum of the squares of the roots of two equations.

Equation 1: $$|x - 2|^2 + |x - 2| - 2 = 0$$

Let $$t = |x - 2|$$ where $$t \geq 0$$. The equation becomes:

$$$t^2 + t - 2 = 0$$$

Factoring:

$$$(t + 2)(t - 1) = 0$$$

So $$t = -2$$ or $$t = 1$$. Since $$t = |x - 2| \geq 0$$, we reject $$t = -2$$.

Therefore $$|x - 2| = 1$$, which gives:

$$x - 2 = 1 \implies x = 3$$ or $$x - 2 = -1 \implies x = 1$$

Sum of squares of roots from Equation 1: $$3^2 + 1^2 = 9 + 1 = 10$$

Equation 2: $$x^2 - 2|x - 3| - 5 = 0$$

Case (a): $$x \geq 3$$

$$|x - 3| = x - 3$$, so the equation becomes:

$$$x^2 - 2(x - 3) - 5 = 0$$$

$$$x^2 - 2x + 6 - 5 = 0$$$

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

$$$(x - 1)^2 = 0 \implies x = 1$$$

But $$x = 1 \lt 3$$, so this does not satisfy our assumption $$x \geq 3$$. No valid root in this case.

Case (b): $$x \lt 3$$

$$|x - 3| = -(x - 3) = 3 - x$$, so the equation becomes:

$$$x^2 - 2(3 - x) - 5 = 0$$$

$$$x^2 + 2x - 6 - 5 = 0$$$

$$$x^2 + 2x - 11 = 0$$$

Using the quadratic formula:

$$$x = \frac{-2 \pm \sqrt{4 + 44}}{2} = \frac{-2 \pm \sqrt{48}}{2} = \frac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3}$$$

Checking: $$x_1 = -1 + 2\sqrt{3} = -1 + 3.464 = 2.464 \lt 3$$ ✓

$$x_2 = -1 - 2\sqrt{3} = -1 - 3.464 = -4.464 \lt 3$$ ✓

Both roots are valid.

Sum of squares of roots from Equation 2:

$$x_1^2 + x_2^2 = (-1 + 2\sqrt{3})^2 + (-1 - 2\sqrt{3})^2$$

Expanding each:

$$(-1 + 2\sqrt{3})^2 = 1 - 4\sqrt{3} + 12 = 13 - 4\sqrt{3}$$

$$(-1 - 2\sqrt{3})^2 = 1 + 4\sqrt{3} + 12 = 13 + 4\sqrt{3}$$

Adding: $$x_1^2 + x_2^2 = (13 - 4\sqrt{3}) + (13 + 4\sqrt{3}) = 26$$

Total sum of squares:

$$$10 + 26 = 36$$$

Hence, the correct answer is Option B.