A plane is formed by the lines through a common point $(1, 1, -1)$ and having direction ratios $2, 1, -3$
and $2, -1, -1$. The distance of the point $(2, 2, 3)$ from the plane is

The normal of plane is perpendicular to the two lines given.

$\therefore$ The plane is $\left|\begin{array}{ccc}x-1& y-1& z+1\\ 2& 1& -3\\ 2& -1& -1\end{array}\right|=0$

$(x-1)(-4)-(y-1)\left(4\right)+(z+1)(-4)=0\Rightarrow x+y+z=1$

The distance of $(2, 2, 3)$ from it is 

$\frac{2+2+3-1}{\sqrt{1+1+1}}=\frac{6}{\sqrt{3}}=2\sqrt{3}$