Answer
$6$
Work Step by Step
Line 1 goes from $(1,2,6)$ and $(2,4,8)$
Direction vector of first line is $v_1=\lt 1,2,2\gt $ and
Direction vector of second line is $v_2=\lt 0,4,2 \gt $
Two skew lines are lying on two parallel planes.
$n=v_1 \times v_2=\lt -4,-2,4\gt $
Use the point $(3,4,0)$ from line 2 and $n$ to form the plane equation.
From the plane equation, we have
$-4x-2y+4z+20=0$
Use points $(1,2,6)$ for line 1.
Distance formula between a point and a plane is given as:
$D=\frac{|ax_1+by_1+cz_1+d|}{\sqrt {a^2+b^2+c^2}}$
$D=\frac{|-4(1)-2(2)+4(6)+20|}{\sqrt {(-4)^2+(-2)^2+4^2}}$
$=\frac{36}{6}$
$=6$
