Let $P Q$ subtend the angle $\theta$ at the origin. $\Rightarrow \angle \mathrm{POQ}=\theta$
Let the slope of line is $\mathrm{m}$.
$\therefore$ The equation of line is $y+2=m(x-1)$
$\Rightarrow \mathrm{mx}-\mathrm{y}=2+\mathrm{m} \Rightarrow \frac{\mathrm{mx}-\mathrm{y}}{2+\mathrm{m}}=1$ ...(i)
The given curve is $3 x^2-y^2-2 x+4 y=0$ ...(ii)
Homogenising eq $^{\mathrm{n}}$ (ii),
we get $3 x^2-y^2-2 x(1)+4 y(1)=0$
$\begin{aligned} & \Rightarrow 3 x^2-y^2-2 x\left(\frac{m x-y}{2+m}\right)+4 y\left(\frac{m x-y}{2+m}\right)=0 \\ & \Rightarrow 3(2+m) x^2-(2+m)-2 m x^2+2 x y+4 m x y-4 y^2=0 \\ & \Rightarrow x^2(6+3 m-2 m)-y^2(2+m+4)+y(4 m+2)=0 \\ & \Rightarrow(6+m) x^2-(6+m) y^2+(4 m+2) x y=0\end{aligned}$
Now, Coefficient of $x^2+$ Coefficient of $y^2$ $=(6+m)-(6+m)=0$
$\therefore$ The angle subtended by $\mathrm{PQ}$ is $90^{\circ}$.