The system of equation has a non-zero solution
$\therefore \begin{vmatrix}a^3&(a+1)^3&(a+2)^3\\a&a+1&(a+2)\\1&1&1\end{vmatrix}=0$
Applying $C_2=C_2-C_1,C_3=C_3-C_2$ we get
$\begin{vmatrix}a^3&3a^2+3a+1&3(a+1)^2+3(a+1)+1\\1&1&1\\1&0&0\end{vmatrix}=0$
$\Rightarrow 3a^2+3a+1-[3(a+1)^2+3(a+1)+1]=0$
$\Rightarrow 3a^2+3a+1-[3(a^2+1+2a)+3a+3+1]=0$
$\Rightarrow 3a^2+3a+1-[3a^2+3+6a+3a+4]=0$
$\Rightarrow 3a^2+3a+1-3a^2-3-6a-3a-4=0$
$\Rightarrow -6a-6=0$
$\Rightarrow -6a=6$
$\Rightarrow a=-1$
Hence (C) is the correct answer.