JEE MAIN - Mathematics (2025 - 2nd April Morning Shift - No. 2)
Let the vertices Q and R of the triangle PQR lie on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}, \mathrm{QR}=5$ and the coordinates of the point $P$ be $(0,2,3)$. If the area of the triangle $P Q R$ is $\frac{m}{n}$ then :
$2 \mathrm{~m}-5 \sqrt{21} \mathrm{n}=0$
$\mathrm{m}-5 \sqrt{21} \mathrm{n}=0$
$5 \mathrm{~m}-21 \sqrt{2} \mathrm{n}=0$
$5 \mathrm{~m}-2 \sqrt{21} \mathrm{n}=0$
Explanation
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$$\begin{aligned} & H:(5 \lambda-3,2 \lambda+1,3 \lambda-4) \\ & <\mathrm{DR} \text { of } \mathrm{PH}> \\ & <5 \lambda-3,2 \lambda-1,3 \lambda-7> \\ & \overrightarrow{P H} \cdot \overrightarrow{Q R}=0 \\ & \Rightarrow(5 \lambda-3) 5+(2 \lambda-1) 2+(3 \lambda-7) 3=0 \\ & \Rightarrow 25 \lambda-15+4 \lambda-2+9 \lambda-21=0 \\ & \Rightarrow 38 \lambda=38 \\ & \Rightarrow \lambda=1 \\ & \therefore \quad H(2,3,-1) \\ & P H=\sqrt{4+1+16}=\sqrt{21} \\ & \therefore \quad \text { Area }=\frac{1}{2} \times P H \text { QR } \\ & \quad=\frac{1}{2} \times \sqrt{21} \times 5=\frac{5 \sqrt{21}}{2}=\frac{m}{n} \\ & 2 m-5 \sqrt{21} n=0 \end{aligned}$$
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