Two lines from the family of lines $(1+2\lambda )x-(1+\lambda )y+1=0$ and the line $x+y=5$ form an equilateral triangle. The equation of the two lines can be 

Given family of lines, $(1+2\lambda )x-(1+\lambda )y+1=0$

$\Rightarrow (x-y+1)+\lambda (2x-y)=0$

$\therefore$Family of lines passing through point of intersection of lines

     $x-y+1=0\text{ and }2x-y=0$

i.e. through the point $(1, 2).$

Since, the family of lines form an equilateral triangle with the line $x+y=5.$

$\begin{array}{l}\therefore y-2=\frac{-1\pm \tan {60}^{\circ }}{1\mp (-1)\tan {60}^{\circ }}(x-1)\\ y-2=\frac{-1\pm \sqrt{3}}{1\mp (-\sqrt{3})}(x-1)\Rightarrow y-2=\frac{-1+\sqrt{3}}{1+\sqrt{3}}(x-1)\end{array}$

  and $y-2=\frac{-1-\sqrt{3}}{1-\sqrt{3}}(x-1)\Rightarrow y-2=(2-\sqrt{3})(x-1)$

and  $y-2=(2+\sqrt{3})(x-1)$