JEE Main 2025 (Online) 2nd April Morning ShiftQuestion 1
Let the vertices $Q$ and $R$ of the triangle $PQR$ lie on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$, $QR=5$ and the coordinates of the point $P$ be $(0,2,3)$. If the area of the triangle $PQR$ is $\frac{m}{n}$ then :
3D Geometry
Solution
Distance of $P$ from the line: $|\vec{AP}\times\vec d|^2/|\vec d|^2=798/38=21$. Area $=\frac12\cdot5\sqrt{21}=\frac{5\sqrt{21}}2=\frac mn$, so $2m-5\sqrt{21}\,n=0$.