Two pipes in a plant room follow the lines x/1 = (y - 1)/2 = (z + 1)/3 and (x - 1)/1 = (y - 3)/k =…

Step 1. The cross product. $\vec{b}_1 \times \vec{b}_2 = (2 \times (-1) - 3k,\ 3 \times 1 - 1 \times (-1),\ 1 \times k - 2 \times 1) = (-2 - 3k,\ 4,\ k - 2)$.

Step 2. The triple product. $(1, 2, 7) \cdot (-2 - 3k, 4, k - 2) = -2 - 3k + 8 + 7k - 14 = 4k - 8$.

Step 3. Coplanar. $4k - 8 = 0$ gives $k = 2$.

Step 4. The joint. Now $\vec{b}_2 = (1, 2, -1)$. The line $x = \lambda$, $y = 1 + 2\lambda$, $z = -1 + 3\lambda$ meets $x = 1 + \mu$, $y = 3 + 2\mu$, $z = 6 - \mu$. The $x$ equation gives $\lambda = 1 + \mu$. The $y$ equation gives $1 + 2 + 2\mu = 3 + 2\mu$, which holds for every $\mu$, so use $z$: $-1 + 3 + 3\mu = 6 - \mu$ gives $\mu = 1$ and $\lambda = 2$.