Let the vertices Q and R of the triangle PQR lie on the line (x+3)/(5)=(y-1)/(2)=(z+4)/(3)…

Question

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 :

Accepted Answer

$$\begin{aligned} & H:(5 \lambda-3,2 \lambda+1,3 \lambda-4) \ & <\mathrm{DR} \text { of } \mathrm{PH}> \ & <5 \lambda-3,2 \lambda-1,3 \lambda-7> \ & \overrightarrow{P H} \cdot \overrightarrow{Q R}=0 \ & \Rightarrow(5 \lambda-3) 5+(2 \lambda-1) 2+(3 \lambda-7) 3=0 \ & \Rightarrow 25 \lambda-15+4 \lambda-2+9 \lambda-21=0 \ & \Rightarrow 38 \lambda=38 \ & \Rightarrow \lambda=1 \ & \therefore \quad H(2,3,-1) \ & P H=\sqrt{4+1+16}=\sqrt{21} \ & \therefore \quad \text { Area }=\frac{1}{2} \times P H \text { QR } \ & \quad=\frac{1}{2} \times \sqrt{21} \times 5=\frac{5 \sqrt{21}}{2}=\frac{m}{n} \ & 2 m-5 \sqrt{21} n=0 \end{aligned}$$

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