If $$a \in R$$ and the equation $$-3(x - [x])^2 + 2(x - [x]) + a^2 = 0$$ (where $$[x]$$ denotes the greatest integer $$\leq x$$) has no integral solution, then all possible values of $$a$$ lie in the interval:

The given equation is

$$-3\,(x-[x])^{2}+2\,(x-[x])+a^{2}=0,$$

where $$[x]$$ is the greatest integer less than or equal to $$x$$. Because in every real number $$x$$ we can separate the integer part and the fractional part, we put

$$x-[x]=\{x\}=t.$$

The fractional part $$t$$ always satisfies $$0\le t<1$$. Substituting $$t$$ in place of $$x-[x]$$ transforms the equation into a quadratic in $$t$$:

$$-3t^{2}+2t+a^{2}=0\qquad\text{with}\qquad 0\le t<1.$$ Now we inspect the possibility that $$x$$ itself is an integer. For an integral $$x$$ we have $$t=x-[x]=0$$, so the left-hand side becomes

$$-3(0)^{2}+2(0)+a^{2}=a^{2}.$$

Therefore the equation is satisfied by an integer $$x$$ iff

$$a^{2}=0\;\Longrightarrow\;a=0.$$

The question asks for the values of $$a$$ for which the equation has no integral solution, so we must have

$$a\neq 0.$$

However, this condition alone does not determine a unique option. Next we notice that, apart from excluding integral solutions, the quadratic should still be solvable for some admissible fractional part $$t$$ (otherwise there would be no real solution at all, making the question purposeless). Thus we look for those $$a$$ for which a root of

$$f(t)=-3t^{2}+2t+a^{2}=0$$

lies in the open interval $$0<t<1.$$ Because $$f(t)$$ is continuous and concave down (its leading coefficient is $$-3<0$$), a root in $$0<t<1$$ will exist exactly when $$f(0)$$ and $$f(1)$$ have opposite signs.

Computing these values one by one, we have

$$f(0)=a^{2},\qquad f(1)=-3(1)^{2}+2(1)+a^{2}=a^{2}-1.$$

To obtain opposite signs we require

$$f(0)>0 \quad\text{and}\quad f(1)<0.$$

The first inequality $$f(0)>0$$ is simply $$a^{2}>0,$$ which we already have because $$a\neq 0$$. The second inequality $$f(1)<0$$ gives

$$a^{2}-1<0\;\Longrightarrow\;a^{2}<1.$$

Combining both,

$$0<a^{2}<1\;\Longrightarrow\;-1<a<0\quad\text{or}\quad 0<a<1.$$

Thus all admissible real values of $$a$$ lie in

$$( -1,0)\cup(0,1).$$

Hence, the correct answer is Option C.