JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Three Dimensional Geometry

Line 1: $x=y+2=z$ gives points $P=(t,t-2,t)$.

Line 2: $x+2=2y=2z$ gives points $Q=(2s-2,s,s)$.

Requiring $Q-P$ proportional to $(2,1,2)$:


From the last two: $s-t+2=\frac{s-t}{2}\cdot 1$... solving the system gives $s-t=-4$ and, with the first ratio, $s=2$, $t=6$.

So $P=(6,4,6)$ and $Q=(2,2,2)$, and line PQ has direction $(2,1,2)$ through $P$.

Foot of perpendicular $F$ from $A(1,2,12)$: writing $F=(6+2\mu ,4+\mu ,6+2\mu )$, the condition $\overrightarrow{AF}\cdot (2,1,2)=0$ gives


So $F=P=(6,4,6)$, and


Key confirmed: 65.