Let the vertices $Q$ and $R$ of the triangle $P Q R$ lie on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}, Q R=5$ 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 :

• A

  $\mathrm{m}-5 \sqrt{21} \mathrm{n}=0$
• Correct answer
  B

  $2 \mathrm{~m}-5 \sqrt{21} \mathrm{n}=0$
• C

  $5 \mathrm{~m}-2 \sqrt{21} \mathrm{n}=0$
• D

  $5 m-21 \sqrt{2} n=0$

Solution

$\mathrm{M}(5 \lambda-3,2 \lambda+1,3 \lambda-4)$
Drs of PM $\Rightarrow 5 \lambda-3,2 \lambda-1,3 \lambda-7$
Drs of line $L \Rightarrow 5,2,3$
$\mathrm{PM} \perp \mathrm{L}$
$\Rightarrow(5 \lambda-3) 5+(2 \lambda-1) 2+(3 \lambda-7) 3=0$
$\Rightarrow \lambda=1$
$\therefore \mathrm{M}(2,3,-1)$
$P M=\sqrt{4+1+16}=\sqrt{21}$
Area $=\frac{1}{2} \times 5 \times \sqrt{21}=\frac{\mathrm{m}}{\mathrm{n}}$
$2 \mathrm{~m}-5 \sqrt{21} \mathrm{n}=0$

Hence, the correct answer is option: 'B'.