Let the vertices Q and R of the triangle P Q R lie on the line (x+3)/(5)=(y-1)/(2)=(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 (m)/(n) then :

Correct answer: B. 2 ~m-5 21 n=0 M(5 λ-3,2 λ+1,3 λ-4) Drs of PM 5 λ-3,2 λ-1,3 λ-7 Drs of line L 5,2,3 PM L (5 λ-3) 5+(2 λ-1) 2+(3 λ-7) 3=0 λ=1 M(2,3,-1) P M= 4+1+16 = 21 Area =(1)/(2) × 5 × 21 = m n 2 ~m-5 21 n=0

JEE Main 20252 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2025 · 2 April · Morning Shift · Mathematics · Three Dimensional Geometry

Let the vertices Q and R of the triangle P Q R lie on the line...

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

(0,2,3)

. If the area of the triangle

P Q R

\frac{m}{n}

then :

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

Options

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

Correct answer

B. $2 \mathrm{~m}-5 \sqrt{21} \mathrm{n}=0$

Step-by-step 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

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