1. Interpret the expression

Given $-3(x-[x])^2+2(x-[x])+a^2=0,$ where $[x]$ is the greatest integer less than or equal to $x$.

Let $t=x-[x].$ Then $t$ is the fractional part of $x$, so $0\le t<1.$

Thus the equation becomes $-3t^2+2t+a^2=0.$

We want this equation to have no integral solution in $x$.

2. Check when $x$ is an integer

If $x$ is an integer, then $x-[x]=0,$ so $t=0$.

Substitute into the equation: $-3(0)^2+2(0)+a^2=0 \implies a^2=0 \implies a=0.$

So:

• if $a=0$, then every integer $x$ is a solution;
• if $a\ne 0$, then there is no integral solution.

Hence all values of $a$ for which there is no integral solution are simply $a\ne 0.$

3. Match with the given options

Now check the options:

• A: $(-2,-1)$ — contained in $a\ne 0$, so these values do give no integral solution.
• B: $(-\infty,-2)\cup(2,\infty)$ — also contained in $a\ne 0$.
• C: $(-1,0)\cup(0,1)$ — also contained in $a\ne 0$.
• D: $(1,2)$ — also contained in $a\ne 0$.

So the actual set of all such $a$ is $\mathbb R\setminus\{0\} = (-\infty,0)\cup(0,\infty).$ This is not equal to any one of the listed intervals.

4. Conclusion

The correct set should be $(-\infty,0)\cup(0,\infty),$ not just option C.

Therefore, the stored answer appears to be incorrect.