A surveyor marks three points on a flat piece of ground, in metres: A(1, 0, 2), B(2, 1, 0) and C(0…

Step 1. Two vectors in the plane. $\overrightarrow{AB} = (1, 1, -2)$ and $\overrightarrow{AC} = (-1, 3, -1)$.

Step 2. A normal. $\overrightarrow{AB} \times \overrightarrow{AC} = (1 \times (-1) - (-2) \times 3,\ (-2) \times (-1) - 1 \times (-1),\ 1 \times 3 - 1 \times (-1)) = (5, 3, 4)$. So $m = 4$.

Step 3. The constant. $k = 5 \times 1 + 3 \times 0 + 4 \times 2 = 13$. The plane is $5x + 3y + 4z = 13$.

Step 4. The check. $B$: $10 + 3 + 0 = 13$. $C$: $0 + 9 + 4 = 13$.

Step 5. The distance. $\dfrac{13}{\sqrt{25 + 9 + 16}} = \dfrac{13}{\sqrt{50}} \approx 1.838$ metres.