Finding difference of angles in triangle . If $\sin A+\sin B+\sin C=0$, $\cos A+\cos B+\cos C=0$, then prove that $$A-B=B-C=C-A=\dfrac{2\pi}3$$

Squaring and adding we get $$\cos(A-B)+\cos(B-C)+\cos(C-A)=-\frac32$$
Again, we have $$\sin A+\sin B=-\sin C\text{ and }\cos A+\cos B=-\cos C$$
Squaring and adding we get $$2+2\cos(A-B)=1\implies \cos(A-B)=-\frac12=\cos\frac{2\pi}3$$
$$\implies A-B=2n\pi\pm\frac{2\pi}3$$ where $n$ is any integer
Case $A:$  Taking the '+' sign,
$$\implies\sin A=\sin\left(B+2n\pi+\frac{2\pi}3\right)=\sin\left(B+\frac{2\pi}3\right)=-\frac12\sin B+\frac{\sqrt3}2\cos B $$
and similarly, $$\cos A=\cos\left(B+2n\pi+\frac{2\pi}3\right)=\cos\left(B+\frac{2\pi}3\right)=-\frac12\cos B-\frac{\sqrt3}2\sin B $$
So, we need $$\sin C=-\frac12\sin B-\frac{\sqrt3}2\cos B\ \  \  \  (1)\text{ and } \cos C=-\frac12\cos B+\frac{\sqrt3}2\cos B\  \   \  \ (2)$$
$$(1)\implies\sin\left(B+\frac\pi3\right)=-\sin C=\sin(-C)\implies B+\frac\pi3=r\pi+(-1)^r(-C) $$
If $r$ is  even $\displaystyle=2m,$(say) $\displaystyle B+\frac\pi3=2m\pi-C\implies B+C=2m\pi-\frac\pi3\  \   \    \ (3)$
If $r$ is  odd $=2m+1,$(say) $\displaystyle B+\frac\pi3=(2m+1)\pi+C\implies B-C=2m\pi+\frac{2\pi}3 \  \   \    \ (4)$
$$(2)\implies\cos\left(B-\frac{2\pi}3\right)=\cos C\implies B+\frac{2\pi}3=2a\pi\pm C $$
Taking the '+' sign, $$ B-\frac{2\pi}3=2a\pi+C\implies B-C=2a\pi+\frac{2\pi}3  \  \   \    \ (5)$$
Taking the '-' sign, $$ B-\frac{2\pi}3=2a\pi-C\implies B+C=2a\pi+\frac{2\pi}3 \  \   \    \ (6)$$
From $(3),(4),(5),(6);$  $$B-C=2a\pi+\frac{2\pi}3 $$ with $A-B=2n\pi+\frac{2\pi}3$ will satisfy the given condition (where $a,n$) are integers
$\displaystyle\implies C-A=-(A-B)-(B-C)$
$\displaystyle=-2n\pi-\frac{2\pi}3-2a\pi-\frac{2\pi}3=2(-a-n-1)\pi+\frac{2\pi}3$
Case $B:$   I leave this as an exercise for you to reach the other condition as Case $A$
