If two members of family $(2+\lambda )x+(1+2\lambda )y-3(1+\lambda )=0$ and line $x+y=0$ make an equilateral triangle, the incentre of triangle is $(a,b)$ , then $\frac{a}{b}=$

Given equation of family of lines is $(2x+y-3)+\lambda (x+2y-3)=0,\lambda \in R$

The family of lines is concurrent at point of intersection of $2x+y-3=0$ and $x+2y-3=0$ which is $A(1,1)$

Two members $L_{1}and{L}_2$ of this family and line $x+y=0$ form an equilateral triangle.
Let foot perpendicular from $A(1,1)$ on the line $x+y=0$ be $M(h,k).$

$\frac{h-1}{1}=\frac{k-1}{1}=\frac{-(1+1)}{2}$
$\therefore M\equiv (h,k)\equiv (0,0)$

Since triangle is equilateral, AM is median, altitude and angle bisector.

$AM=\sqrt{{(1-0)}^2+{(1-0)}^2}=\sqrt{2}$

Incentre coincides with centroid G such that $AG=\frac{2}{3}\sqrt{2}.$

Slope of AM is $1=\tan {45}^0$

$\therefore G=(1-\frac{2\sqrt{2}}{3}\cos {45}^0,1-\frac{2\sqrt{2}}{3}\sin {45}^0)\equiv (\frac{1}{3},\frac{1}{3})$