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QJEEM25Ma5d44: JEE Main 2025 Mathematics - Class 12 JEEM25Ma5d44 Solution: Let P be the image of the point in the line and be a point on . …

Let P be the image of the point Q(7,−2,5)\mathrm{Q}(7,-2,5) in the line L:x−12=y+13=z4\mathrm{L}: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4} and R(5,p,q)\mathrm{R}(5, \mathrm{p}, \mathrm{q}) be a point on LL . Then the square of the area of △PQR\triangle P Q R is _________.

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JEE Main 2025 (Online) 24th January Evening Shift Mathematics - 3D Geometry Question 3 English Explanation

amp; Let R(2λ+1,3λ−1,4λ)amp;amp;2λ+1=5amp;λ=2amp;R(5,5,8)amp; let T(2λ+1,3λ−1,4λ)amp;QT→=(2λ−6)i^+(3λ+1)j^+(4λ−5)k^amp; b→=2i^+3j^+4k^amp;QT→⋅ b→=0amp;4λ−12+9λ+3+16λ−20=0\begin{aligned}&\text { Let } \mathrm{R}(2 \lambda+1,3 \lambda-1,4 \lambda)\\&\begin{aligned}& 2 \lambda+1=5 \\& \lambda=2 \\& \mathrm{R}(5,5,8) \\& \text { let } \mathrm{T}(2 \lambda+1,3 \lambda-1,4 \lambda) \\& \overrightarrow{\mathrm{QT}}=(2 \lambda-6) \hat{\mathrm{i}}+(3 \lambda+1) \hat{\mathrm{j}}+(4 \lambda-5) \hat{\mathrm{k}} \\& \overrightarrow{\mathrm{~b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\& \overrightarrow{\mathrm{QT}} \cdot \overrightarrow{\mathrm{~b}}=0\end{aligned}\\&4 \lambda-12+9 \lambda+3+16 \lambda-20=0\end{aligned}

amp;λ=1amp; T(3,2,4)amp;QT=33RT=29amp;( area of ΔPQR)2=(1229⋅233)2amp;=957\begin{aligned}& \lambda=1 \\& \mathrm{~T}(3,2,4) \\& \mathrm{QT}=\sqrt{33} \quad \mathrm{RT}=\sqrt{29} \\& (\text { area of } \Delta \mathrm{PQR})^2=\left(\frac{1}{2} \sqrt{29} \cdot 2 \sqrt{33}\right)^2 \\& =957\end{aligned}

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