The correct option is D (−2,3)
Let z=α be a real root. Then,
α3+(3+2i)α+(−1+ia)=0
⇒(α3+3α−1)+i(a+2α)=0
⇒α3+3α−1=0 and α=−a2
⇒−a38−3a2−1=0
⇒a3+12a+8=0
Let f(a)=a3+12a+8\
∴f(−1)<0,f(0)>0,f(−2)<0,f(1)>0
and f(3)>0.
Hence, a∈(−1,0) or a∈(−2,1) or a∈(−2,3)