Two planes x + 2y - z = 1 and 2x - y + z = 4 meet in a line, with lengths in metres…

Step 1. The condition. $3(1 + 2\lambda) + (2 - \lambda) = 5 + 5\lambda = 0$, so $\lambda = -1$.

Step 2. The plate. $(x + 2y - z - 1) - (2x - y + z - 4) = -x + 3y - 2z + 3 = 0$. Multiply by $-1$: $x - 3y + 2z = 3$. So $m = -3$ and $k = 3$.

Step 3. The distance. The normal $(1, -3, 2)$ has length $\sqrt{14}$. The distance is $\dfrac{3}{\sqrt{14}} \approx 0.802$ metres.