Two straight pipes in a chemical plant pass without touching…

Step 1. The vector $\overrightarrow{PQ}$. $Q - P = (1 + \mu - \lambda,\ -6 - \mu + 2\lambda,\ 1 - 2\lambda)$.

Step 2. Perpendicular to $\vec{b}_1 = (1, -2, 2)$. $(1 + \mu - \lambda) - 2(-6 - \mu + 2\lambda) + 2(1 - 2\lambda) = 0$ gives $15 + 3\mu - 9\lambda = 0$, so $\mu = 3\lambda - 5$.

Step 3. Perpendicular to $\vec{b}_2 = (1, -1, 0)$. $(1 + \mu - \lambda) - (-6 - \mu + 2\lambda) = 0$ gives $7 + 2\mu - 3\lambda = 0$.

Step 4. Solve. Put $\mu = 3\lambda - 5$ into the second equation: $7 + 6\lambda - 10 - 3\lambda = 0$, so $\lambda = 1$ and $\mu = -2$.

Step 5. The feet. $P = (-1, 1, 4)$ and $Q = (-3, -1, 3)$. The $z$-coordinate of $P$ is $4$.

Step 6. The length. $PQ = \sqrt{4 + 4 + 1} = 3$ metres.

Step 7. The check. $\vec{b}_1 \times \vec{b}_2 = (-2 \times 0 - 2 \times (-1),\ 2 \times 1 - 1 \times 0,\ 1 \times (-1) - (-2) \times 1) = (2, 2, 1)$ with length $3$. The gap is $(1, -6, 1)$, and $(1, -6, 1) \cdot (2, 2, 1) = -9$. So $d = \dfrac{9}{3} = 3$ metres. The two methods agree.