Let the minimum value $$v_0$$ of $$v=|z|^{2} + |z-3|^{2} + |z-6i|^{2}, z ∈ \mathbb{C} $$ s attained at $$z=z_{0}$$. Then $$|2z_0^2- \overline{z}_0^3+3|^{2}+v_0^2 $$ is equal to

We have $$f(x) = \displaystyle\sum_{k=1}^{13}(x-k)^2$$. To find the minimum, we take the derivative and set it to zero.

$$f'(x) = 2\displaystyle\sum_{k=1}^{13}(x-k) = 2\left(13x - \sum_{k=1}^{13}k\right) = 2\left(13x - \dfrac{13 \times 14}{2}\right) = 2(13x - 91)$$

Setting $$f'(x) = 0$$: $$13x = 91$$, so $$x_0 = 7$$.

Since $$f''(x) = 26 > 0$$, this is indeed a minimum.

Now we compute $$\binom{13}{x_0} = \binom{13}{7} = \dfrac{13!}{7! \cdot 6!}$$.

$$= \dfrac{13 \times 12 \times 11 \times 10 \times 9 \times 8}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = \dfrac{1235520}{720} = 1716$$

The answer is Option A: $$1716$$.