Answer
Skew, $\frac{1}{\sqrt 6}$
Work Step by Step
Direction vector of the first line is$v_1=\lt 1,1,1\gt $ and the direction vector of the second line is $v_1=\lt 1,2,3\gt $, so the lines are not parallel because the components are not proportional.
For line 1: $x=t,y=t,z=t$
For line 2: $x=-1+s,y=2s,z=3s$
Set the corresponding $x,y,z$ equations equals to each other.
$s=-1$ and $t=2(-1)=-2$ and $-2\ne 3(-1) = -3$
Thus, there is no solution so the lines do not intersect. Hence, they are skew.
Distance formula between a point and a plane is given as:
$D=\frac{|ax_1+by_1+cz_1+d|}{\sqrt {a^2+b^2+c^2}}$
$D=\frac{|1(0)-2(0)+1(0)+1|}{\sqrt {1^2+(-2)^2+1^2}}$
$=\frac{1}{\sqrt 6}$
