Two drones fly along straight paths over a wide field…

Step 1. The crossing point. Equal coordinates give $1 + 3s = 5 + u$, $3 + 3s = 7 + u$ and $10 + 2s = 6 + 2u$. The first two are the same equation, $u = 3s - 4$. The third gives $10 + 2s = 6 + 6s - 8$, so $4s = 12$ and $s = 3$. Then $u = 5$, and the point is $(10, 12, 16)$.

Step 2. The times. Drone A moves with $t = s$, so it reaches the point at $t = 3$ minutes. Drone B moves with $t = u$, so it reaches it at $t = 5$ minutes. The drones are at the crossing point at different times, so they do not collide there.

Step 3. The separation. The vector from B to A at time $t$ is $(-4 + 2t,\ -4 + 2t,\ 4)$. Its squared length is $2(2t - 4)^2 + 16$.

Step 4. The minimum. This is smallest when $2t - 4 = 0$, so $t = 2$ minutes. The squared length is then $16$ and the distance is $4$ hundred metres, which is $400$ metres.