Let $$f(x) = \begin{vmatrix} a & -1 & 0 \\ ax & a & -1 \\ ax^2 & ax & a \end{vmatrix}, a \in R$$. Then the sum of the squares of all the values of $$a$$ for
$$2f'(10) - f'(5) + 100 = 0$$ is

We are given $$f(x) = \begin{vmatrix} a & -1 & 0 \\ ax & a & -1 \\ ax^2 & ax & a \end{vmatrix}$$ and need to find the sum of squares of all values of $$a$$ for which $$2f'(10) - f'(5) + 100 = 0$$.

Expanding along the first row, $$f(x) = a(a^2 + ax) - (-1)(a^2x + ax^2) + 0 = a(a^2 + ax) + (a^2x + ax^2) = a^3 + a^2x + a^2x + ax^2 = a^3 + 2a^2x + ax^2 = a(a^2 + 2ax + x^2) = a(a + x)^2$$, so $$f(x) = a(a + x)^2$$.

From this, $$f'(x) = a \cdot 2(a + x) = 2a(a + x)$$, giving $$f'(10) = 2a(a + 10)$$ and $$f'(5) = 2a(a + 5)$$.

Substituting into $$2f'(10) - f'(5) + 100 = 0$$ leads to $$2 \cdot 2a(a + 10) - 2a(a + 5) + 100 = 0$$, which simplifies to $$4a(a + 10) - 2a(a + 5) + 100 = 0$$, then to $$4a^2 + 40a - 2a^2 - 10a + 100 = 0$$, and finally to $$2a^2 + 30a + 100 = 0$$ or $$a^2 + 15a + 50 = 0$$.

Solving $$(a + 5)(a + 10) = 0$$ gives $$a = -5$$ or $$a = -10$$, and hence the sum of squares is $$(-5)^2 + (-10)^2 = 25 + 100 = 125$$. Therefore, the sum of squares of all values of $$a$$ is 125. The correct answer is Option C: $$125$$.