#### Answer

$\frac{13}{\sqrt {69}}$

#### Work Step by Step

Line 1 goes from the origin and $(2,0,-1)$
Line 1 goes from $(1,-1,1)$ and $(4,1,3)$
Direction vector of first line is $v_1=\lt 2,0,-1\gt $ and
Direction vector of second line is $v_2=\lt 3,2,2 \gt $
Two skew lines are lying on two parallel planes.
$n=v_1 \times v_2=\lt 2,-7,4\gt $
Use the point $(1,-1,1)$ from line 2 and $n$ to form the plane equation.
From the plane equation, we have
$2x-7y+4z-13=0$
Use point $(0,0,0)$ for line 1.
Distance formula between a point and a plane is given as:
$D=\frac{|ax_1+by_1+cz_1+d|}{\sqrt {a^2+b^2+c^2}}$
$D=\frac{|2(0)-7(0)+4(0)-13|}{\sqrt {2^2+(-7)^2+4^2}}$
$=\frac{13}{\sqrt {69}}$