Complexity of $a$ , $b$ , and $c$

Consider $b$ , $c$ and $-a$ as vectors added together in an Argand diagram:

All the vectors shown above have a length of $1$ , as $|a|=|b|=|c|=|b+c-a|=1$ . Therefore, they must be the sides of a rhombus. Since opposite sides of a rhombus are parallel and of equal length, and the $-a$ vector is both negative and pointing in the opposite direction to the $b$ vector, we can see that $-(-a)=b$ , therefore $a=b$ .

However, the question states that the complex numbers $a$ , $b$ and $c$ are all distinct. Therefore $a≠b$ , and the rhombus above cannot be a true rhombus. However, if $b+c=0$ , then $a$ can be any value where $|a|=1$ :

So the answer is: $\begin{aligned}3-|b+c|&=3-0\\&=\boxed{3}\end{aligned}$