$$
\begin{aligned}
& \text { Let } y=\frac{a x^2-2 x+3}{2 x-3 x^2+a} \\
& \Rightarrow \quad 2 x y-3 x^2 y+a y=a x^2-2 x+3 \\
& \Rightarrow \quad x^2(a+3 y)-2 x(y+1)+3-a y=0 \\
& \text { as } x \in R, D \geq 0 \\
& \therefore \quad 4(y+1)^2-4(a+3 y)(3-a y) \geq 0
\end{aligned}
$$

$$
\begin{aligned}
& \Rightarrow \quad\left(y^2+2 y+1\right)-\left[3 a-a^2 y+9 y-3 a y^2\right) \geq 0 \\
& \Rightarrow \quad y^2(3 a+1)+\left(a^2-7\right) y+1-3 a \geq 0 \\
& \text { as } y \in R, D \leq 0 \\
& \therefore \quad\left(a^2-7\right)^2+4(3 a+1)(3 a-1) \leq 0 \\
& \Rightarrow a^4-14 a^2+49+36 a^2-4 \leq 0 \\
& \Rightarrow a^4+22 a^2+45 \leq 0 \\
& \quad\left[\because a^4+22 a^2+45>0, \forall a \in R\right] \\
& \Rightarrow a \in \phi
\end{aligned}
$$