Answer
$2$
Work Step by Step
For line 1: $x=1+t,y=1+6t,z=2t$
For line 2: $x=1+2s,y=5+15s,z=-2+6s$
Direction vector of first line is $v_1=\lt 1,6,2\gt $ and
Direction vector of second line is $v_1=\lt 2,15,6\gt $
Two skew lines are lying on two parallel planes.
$n=v_1 \times v_2=\lt 6,-2,3\gt $
Letting $s=0$ for line 2: $x=1+2s,y=5+15s,z=-2+6s$, we get the point $(1,5,-2)$
From the plane equation, we have
$6x-2y+3z+10=0$
Now, letting $t=1$ for line 1, we get the point $(1,1,0)$
Distance formula between a point and a plane is given as:
$D=\frac{|ax_1+by_1+cz_1+d|}{\sqrt {a^2+b^2+c^2}}$
$D=\frac{|6(1)-2(1)+3(0)+10|}{\sqrt {6^2+(-2)^2+3^2}}$
$=\frac{14}{7}$
$=2$
