2019 AMC 12B Problem 24

As $a,b,c$ range over $[0,1],$ the set $S$ is the Minkowski sum of the three unit segments along $v_1=1=(1,0),$ $\ v_2=\omega=\left(-\dfrac12,\dfrac{\sqrt3}{2}\right),$ $\ v_3=\omega^2=\left(-\dfrac12,-\dfrac{\sqrt3}{2}\right).$ This is a zonogon whose area is the sum of the cross-product magnitudes over pairs. Each pair gives $|v_i\times v_j|=\dfrac{\sqrt3}{2}.$ Therefore the area is $3\cdot\dfrac{\sqrt3}{2}=\dfrac{3\sqrt3}{2}.$ Thus, C is the correct answer.