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 equal to $\frac{k}{13}\left(3\sqrt{3}-1\right)$  then value of k is____

The abscissa of the intersection points of the given line and given curve is.

$x^3+{\left(\sqrt{3}x\right)}^3+3x\left(\sqrt{3}x\right)+5x^2$$+3{\left(\sqrt{3}x\right)}^2+4x+5\left(\sqrt{3}x\right)-1=0$

$\therefore \left(3\sqrt{3}+1\right)x^3+\left(3\sqrt{3}+14\right)x^2$$+\left(5\sqrt{3}+4\right)x-1=0$

$\therefore x_1{x}_2{x}_3=\frac{1}{3\sqrt{3}+1}$

Hence  $OA.OB.OC=2x_1.2x_2.2x_3$

$=8.x_1{x}_2{x}_3$

$=8.\frac{1}{\sqrt{3}+1}\times \frac{3\sqrt{3}-1}{3\sqrt{3}-1}$

$=\frac{8}{26}.\left(3\sqrt{3}-1\right)=\frac{4}{13}\left(3\sqrt{3}-1\right)$

$\therefore k=4$