Solution
= Solution
The cross product of the plane normals is proportional to $(1,-1,-1)$, so
$$
m=\frac1{\sqrt3}(1,-1,-1).
$$
The point on the intersection perpendicular to $m$ is
$$
u=\frac19(11,4,7),
$$
which satisfies both plane equations and $u\cdot m=0$. Hence the line is $r\times m=u\times m$.
The equally inclined line has direction $d=(1,1,1)/\sqrt3$. The distance between the two skew lines is
$$
\boxed{\frac{|u\cdot(m\times d)|}{|m\times d|}
=\frac1{3\sqrt2}.}
$$