Distinguish a geometric flight path from motion in time

Write $\mathbf r=OA+t(-2,1,1)$. At t=6, $(-9,4,6)=OA+6(-2,1,1)$, so $OA=(3,-2,0)$. Departure is A=(3,−2,0), consistent with ground level.

$OF=(3,-2,0)+t(-2,1,1)$ and $CF=(10-2t,-7+t,t-2)$. A perpendicular foot satisfies $CF\cdot(-2,1,1)=-20+4t-7+t+t-2=0$. Thus $t=29/6$≈4.83333≥0, so the foot is on the flown ray. Then $CF=(1/3,-13/6,17/6)$ and $|CF|=\sqrt{(1/3)^2+(-13/6)^2+(17/6)^2}=\sqrt{77/6}=3.58$m(3s.f.).

$AC=OC-OA=(-10,7,2)$. The distance is $|AC\times(-2,1,1)|/|(-2,1,1)|=|(5,6,4)|/\sqrt6=\sqrt{77}/\sqrt6=3.58$m(3s.f.). The valid positive foot parameter from method 1 ensures this line distance is also the ray distance.

$|AF|=|AC\cdot(-2,1,1)|/\sqrt6=29/\sqrt6$. Thus distance=$\sqrt{|AC|^2-|AF|^2}=\sqrt{(-10)^2+7^2+2^2-(29/\sqrt6)^2}=\sqrt{77/6}=3.58$m(3s.f.).

Set the common ratio to λ: $x=1/2-3\lambda$, $y=-7+4\lambda$, $z=10+k\lambda$. Hence the requested path representation is $\mathbf r=(1/2,-7,10)+\lambda(-3,4,k)$, λ≥0. Thus $\mathbf p=(1/2,-7,10)$ and $\mathbf q=(-3,4,k)$.

At an intersection, $(1/2,-7,10)+\lambda(-3,4,k)=(3,-2,0)+t(-2,1,1)$. Coordinates give $-3\lambda+2t=5/2$, $4\lambda-t=5$, $10+k\lambda=t$. The first two yield λ=2.5 and t=5; the third gives $10+2.5k=5$, so k=−2. Both parameters are nonnegative and the intersection is (−7,3,5).

The printed solution says “No collision” because λ=2.5 and t=5 are different arrival times. That conclusion requires the extra assumption that λ is elapsed time in seconds from a drone starting at p at takeoff, with velocity q. The prompt specifies only a steady-speed path for the drone, not that time parameterisation. Geometric parameters need not equal time. If λ(t)=t, it does not collide; if λ(t)=t/2 from the same starting point, it reaches the intersection at t=5 and does collide. Both motions have steady speed. Therefore the collision cannot be determined from the stated path and simultaneous takeoff alone; drone velocity or an explicit λ=time condition is needed.