JEE Main 2022 (Online) 26th July Morning ShiftQuestion 1
Let $Q$ and $R$ be two points on the line $\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}$ at a distance $\sqrt{26}$ from the point $P(4,2,7)$. Then the square of the area of the triangle $PQR$ is ________.
3D Geometry
Answer$153$
Solution
Point on line $X=(-1+2t,-2+3t,1+2t)$. $|XP|^2=(2t-5)^2+(3t-4)^2+(2t-6)^2=17t^2-64t+77=26\Rightarrow17t^2-64t+51=0\Rightarrow t=1,3$. $Q=(1,1,3)$, $R=(5,7,7)$. $\vec{PQ}=(-3,-1,-4)$, $\vec{PR}=(1,5,0)$. Cross product $=(20,-4,-14)$, squared magnitude $=400+16+196=612$. Area$^2=612/4=153$.