1. Let $w_k=z_k-z_0 \\ (k=1,2,3).$ Then $w_1,w_2,w_3$ are the position vectors of the vertices of the equilateral triangle with centroid shifted to the origin.

2. Since $z_0$ is the centroid, $w_1+w_2+w_3=(z_1-z_0)+(z_2-z_0)+(z_3-z_0)=0.$

3. For an equilateral triangle centered at the origin, the three vertices are equally spaced by angle $\frac{2\pi}{3}$. So for some complex number $a$, $w_1=a, \quad w_2=a\omega, \quad w_3=a\omega^2,$ where $\omega=e^{2\pi i/3}, \qquad \omega^3=1, \qquad 1+\omega+\omega^2=0.$

4. Now compute the required sum: $\sum_{k=1}^3 (z_k-z_0)^2=w_1^2+w_2^2+w_3^2.$ Substitute the above forms: $w_1^2+w_2^2+w_3^2=a^2+a^2\omega^2+a^2\omega^4.$

5. Since $\omega^3=1$, we have $\omega^4=\omega$. Hence $a^2(1+\omega^2+\omega)=a^2(1+\omega+\omega^2)=a^2\cdot 0=0.$

6. Therefore, $\sum_{k=1}^3 (z_k-z_0)^2=0.$

7. Checking options:

   • A: $0$ ✅
   • B: $1$ ❌
   • C: $i$ ❌
   • D: $-i$ ❌

Hence the correct answer is A.