$z^2+az+12=0$

$z_1+z_2=-a$ and $z_1{z}_2=12$

If $0, z_1, z_2$ are vertices of equilateral triangles

$z_2=z_1{e}^{\frac{i\pi }{3}}$

$z_2=z_1\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)$

$2z_2-z_1=\sqrt{3}i{z}_1$

squaring both sides

$\Rightarrow 4z_2^2+z_1^2-4z_1{z}_2=-3z_1^2$

$\Rightarrow {\left(z_1+z_2\right)}^2=3z_1{z}_2$

$\Rightarrow {\mathrm{a}}^2=3\times 12$

$\Rightarrow \left|a\right|=6$