Let ${\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{xy}+1\ge \mathrm{a}(\mathrm{x}+\mathrm{y})\mathrm{\forall }\mathrm{x},\mathrm{y}\in \mathrm{R}$, then the number of possible integer(s) in the range of a is________.

$$\begin{gathered} {\mathrm{x}}^2+\mathrm{x}(\mathrm{y}-\mathrm{a})+{\mathrm{y}}^2-\mathrm{ay}+1\ge 0\mathrm{x}\in \mathrm{R} \\ \begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\mathrm{y}-\mathrm{a}{)}^2-4\left({\mathrm{y}}^2-\mathrm{ay}+1\right)\le 0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-3{\mathrm{y}}^2+2\mathrm{ay}+{\mathrm{a}}^2-4\le 0\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3{\mathrm{y}}^2-2\mathrm{ay}+4-{\mathrm{a}}^2\ge 0\mathrm{y}\in \mathrm{R}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{D}\le 0\end{array} \end{gathered}$$
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4{\mathrm{a}}^2-4,3\left(4-{\mathrm{a}}^2\right)\le 0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}^2-3\left(4-{\mathrm{a}}^2\right)\le 0\Rightarrow 4{\mathrm{a}}^2-12\le 0\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ Range of }\mathrm{a}\in (-\sqrt{3},\sqrt{3})\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ number of integer }\{-1,0,1\}\end{array}$