If lines $x=y=z$ and $x=\frac{y}{2}=\frac{z}{3}$ and third line passing through (1,1,1) form a triangle of area $\sqrt{6}$units, the point of intersection of third lines with second line is _________

$x=y=z\to 1, x=\frac{y}{2}=\frac{z}{3}$

Point of interaction of  1 & 2 is (0,0,0) angle between 1 & 2 is $\sin \theta =\frac{\sqrt{6}}{\sqrt{42}}$ third line passing through (1,1,1)  and satisfying 1 Hence A(1,1,1) 

$Area=\frac{1}{2}\left(OA\right)\left(OB\mathrm{Sin}\theta \right)$

$\sqrt{6}=\frac{1}{2}\sqrt{3}.\sqrt{14}\lambda .\frac{\sqrt{6}}{\sqrt{42}}$

$\lambda =2.$