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

$\begin{array}{l}GivenEquation-3{(x-\left[x\right])}^2+2(x-\left[x\right])+a^2=0\\ Letx-\left[x\right]=y\Rightarrow 0<y<1\\ \Rightarrow 3y^2-2y-a^2=0\\ y=\frac{2\pm \sqrt{4+12a^2}}{6}\Rightarrow \frac{1\pm \sqrt{1+3a^2}}{3}\\ \therefore 0<\frac{1\pm \sqrt{1+3a^2}}{3}<1\\ Let\frac{1\pm \sqrt{1+3a^2}}{3}<1\Rightarrow \sqrt{1+3a^2}<2\\ \Rightarrow 1+3a^2<4\\ \Rightarrow 3a^2-3<0\\ \Rightarrow a^2-1<0\\ \Rightarrow (a+1)(a-1)<0\\ \therefore a\in (-1,1)\\ fora=0,y=0ifisnotpossible\\ \therefore a\in (-1,0)\cup (0,1)\end{array}$