The condition $a^2+b^2+c^2-ab-bc-ca=0$ geometrically implies that $a,b,c$ are vertices of an equilateral triangle (proof at bottom of this answer). It follows that
$$|a-b|=|b-c|=|c-a|=|a|$$
Just as you obtained from $|a-b|=|a|$, $a\bar{b}+\bar{a}b=|b|^2$, combining other two with $|a|$ will give respectively $$a\bar{c}+\bar{a}c=|c|^2$$ and
$$b\bar{c}+\bar{b}c=|b|^2+|c|^2-|a|^2$$
Using these in $|a+b+c|^2=21^2$, $|b|^2+|c|^2$ will be found.
Answer : $|b|^2+|c|^2=21^2/3=147$
Statement : Complex numbers $a,b,c$ are vertices of an equilateral triangle $\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0$.
Proof : I prove $\Leftarrow$ direction. $a^2+b^2+c^2-ab-bc-ca=$$(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)=0$.
So either $(a+b\omega+c\omega^2)=0$ or $(a+b\omega^2+c\omega)=0$
(A) $a+b\omega+c\omega^2=0 \Rightarrow a+b\omega-c-c\omega=0 $ $$\frac{a-c}{b-c}=-\omega = e^{-i\pi/3}\Rightarrow (a-c)=(b-c)e^{-i\pi/3}$$
meaning $a,b,c$ form the vertices of a positively oriented equilateral triangle i.e., $\triangle ABC$ has vertices $A,B,C$ in anticlockwise sense.
(B) $a+b\omega^2+c\omega=0 \Rightarrow a+b\omega^2-c-c\omega^2=0 $ $$\frac{a-c}{b-c}=-\omega^2 = e^{+i\pi/3}\Rightarrow (a-c)=(b-c)e^{+i\pi/3}$$
meaning $a,b,c$ form the vertices of a negatively oriented equilateral triangle i.e., $\triangle ABC$ has vertices $A,B,C$ in clockwise sense. $\quad \blacksquare$
