Let z _ 1 , z _ 2 be the roots of the equation z ^ 2 + az + 12 = 0...

${z}^{2} + az + 12 = 0$

${z}_{1} + {z}_{2} = - a$ and ${z}_{1}{z}_{2} = 12$

If $0, {z}_{1}, {z}_{2}$ are vertices of equilateral triangles

${z}_{2} = {z}_{1}e\dfrac{i\pi}{3}$

${z}_{2} = {z}_{1}\left(\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}\right)$

$2{z}_{2} - {z}_{1} = \sqrt{3}i{z}_{1}$

squaring both sides

$⇒ 4{z}_{2}^{2} + {z}_{1}^{2} - 4{z}_{1}{z}_{2} = - 3{z}_{1}^{2}$

$⇒ {\left({z}_{1} + {z}_{2}\right)}^{2} = 3{z}_{1}{z}_{2}$

$⇒ {a}^{2} = 3 \times 12$

$⇒ \left|a\right| = 6$