JEE Main 2025 Apr 2 Shift 1, Mathematics Q17: Lines in Space

Setup. Take $M$ as the foot of perpendicular from $P$ to the line $L$ containing $QR$. Then $PM$ is the height of $△PQR$, and since $QR$ is a chord on $L$, the foot $M$ is the midpoint only when $PQ=PR$ but the area only needs $PM$ and $QR$.

Parametrise the line as $M(5\lambda -3,2\lambda +1,3\lambda -4)$.

Direction ratios: $\overrightarrow{PM}=(5\lambda -3,2\lambda -1,3\lambda -7)$; line direction $\vec{d}=(5,2,3)$.

Impose $\overrightarrow{PM}\cdot \vec{d}=0$:

$$5(5\lambda -3)+2(2\lambda -1)+3(3\lambda -7)=0$$

$$25\lambda -15+4\lambda -2+9\lambda -21=0$$

$$38\lambda -38=0\Rightarrow \lambda =1$$

Then $M=(2,3,-1)$ and:

$$PM=\sqrt{(2-0{)}^2+(3-2{)}^2+(-1-3{)}^2}=\sqrt{4+1+16}=\sqrt{21}$$

Area of $△PQR$:

$$\text{Area}=\frac{1}{2}\cdot QR\cdot PM=\frac{1}{2}\cdot 5\cdot \sqrt{21}=\frac{5\sqrt{21}}{2}=\frac{m}{n}$$

Hence $\frac{m}{n}=\frac{5\sqrt{21}}{2}$, so $2m=5\sqrt{21}n$, i.e. $2m-5\sqrt{21}n=0$.

$\therefore$ Option (2) is correct.