If the line $y=\sqrt{3}x$ cuts the curve $x^3+y^3+3xy+5x^2+3y^2+4x+5y-1=0$  at the points  $A,B,C,$then $OA.OB.OC$ is

$\mathrm{Tan}\theta =\sqrt{3}\text{ Line is }\frac{x-0}{\frac{1}{2}}=\frac{y-0}{\frac{\sqrt{3}}{2}}=r$

any point on the line is $\left(\frac{r}{2},\frac{\sqrt{3}}{2}r\right)$

Where r is distance from (0,0) substituting in the curve $r^3\left(\frac{1+3\sqrt{3}}{8}\right)+r^2(\dots )+r(\dots .)-1=0$

this is cubic in ‘r’

$\therefore OA\cdot OB.OC=\frac{1}{(1+3\sqrt{3})}\left(8\right)=\frac{4}{13}(3\sqrt{3}-1)$