If the tangent at a point P on the parabola ${\mathrm{y}}^2=3\mathrm{x}$  is parallel to the line $\mathrm{x}+2\mathrm{y}=1$ and the
tangents at the points Q and R on the ellipse $\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{1}=1$ are perpendicular to the line  $x-y=2$ then the area of the triangle PQR is

Slope of the tangent at P(t) on the parabola $y^2=3x$is parallel to the line $\mathrm{x}+2\mathrm{y}=1$is  $\frac{1}{t}=\frac{-1}{2}$$\Rightarrow P=\left(a{t}^2,2at\right)=(3,-3)$
Slope the tangents at the points Q and R on the ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$are perpendicular to the
line $x-y=2\text{ is }m=-1\Rightarrow c^2=a^2{m}^2+b^2\Rightarrow c=\pm \sqrt{5}$
$\Rightarrow Q=\left(\frac{4}{\sqrt{5}},\frac{1}{\sqrt{5}}\right)\text{ and }Q=\left(\frac{-4}{\sqrt{5}},\frac{-1}{\sqrt{5}}\right)$
Area of $\mathrm{△}PQR=3\sqrt{5}\text{ Squints. }$