Connect a cross-product relation to areas and a nonzero linear combination

Because b is a unit vector, $|\mathbf b\cdot\mathbf c|=|\mathbf c\cdot\mathbf b|$ is the nonnegative length of the projection of c onto b. The absolute-value bars in the printed question are essential.

The given vector relation implies $2|\mathbf a\times\mathbf b|=|\mathbf b\times\mathbf c|$. Both triangle areas are half the corresponding cross-product magnitude, so their ratio is $1:2$. The areas are nonzero: if both cross products vanished, a and c would both be parallel to nonzero b, contrary to their 60°angle.

Use antisymmetry to rewrite

$\begin{aligned}2\mathbf a\times\mathbf b-\mathbf b\times\mathbf c&=\mathbf 0,\\2\mathbf a\times\mathbf b+\mathbf c\times\mathbf b&=\mathbf 0,\\(2\mathbf a+\mathbf c)\times\mathbf b&=\mathbf 0.\end{aligned}$

Since b is nonzero, $2\mathbf a+\mathbf c=k\mathbf b$ for real k. This combination cannot be zero: $|2\mathbf a+\mathbf c|\ge 2|\mathbf a|-|\mathbf c|=4>0$. Thus $k\ne 0$.

$\begin{aligned}(2\mathbf a+\mathbf c)\cdot(2\mathbf a+\mathbf c)&=k^2\mathbf b\cdot\mathbf b,\\4|\mathbf a|^2+4\mathbf a\cdot\mathbf c+|\mathbf c|^2&=k^2,\\36+4(3)(2)\cos 60^\circ+4&=k^2,\\52&=k^2.\end{aligned}$

Hence $k=\pm 2\sqrt{13}$. Both signs are retained because the statement does not fix the orientation of b.