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QJEEM25Made59: JEE Main 2025 Mathematics - Class 12 JEEM25Made59 Solution: Let in a , the length of the side be 6 , the vertex be and the…

Let in a △ABC\triangle A B C , the length of the side ACA C be 6 , the vertex BB be (1,2,3)(1,2,3) and the vertices A,CA, C lie on the line x−63=y−72=z−7−2\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2} . Then the area (in sq. units) of △ABC\triangle A B C is:

Multiple Choice Options

  • A 42
  • B 17
  • C 56
  • D 21

Step-by-Step Verified Solution

JEE Main 2025 (Online) 24th January Morning Shift Mathematics - 3D Geometry Question 5 English Explanation

amp; Let M(3λ+6,2λ+7,−2λ+7)amp;BM→=(3λ+5)i^+(2λ+5)j^+(−2λ+4)k^amp;AC→⋅BM→=0=3(3λ+5)+2(2λ+5)−2(−2λ+4)amp;BM→=2i^+3j^+6k^amp;∣BM→=7amp; Area =12×6×7=21\begin{aligned}& \text { Let } \mathrm{M}(3 \lambda+6,2 \lambda+7,-2 \lambda+7) \\& \overrightarrow{\mathrm{BM}}=(3 \lambda+5) \hat{\mathrm{i}}+(2 \lambda+5) \hat{\mathrm{j}}+(-2 \lambda+4) \hat{\mathrm{k}} \\& \overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BM}}=0=3(3 \lambda+5)+2(2 \lambda+5)-2(-2 \lambda+4) \\& \overrightarrow{\mathrm{BM}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} \\& \mid \overrightarrow{\mathrm{BM}}=7 \\& \text { Area }=\frac{1}{2} \times 6 \times 7=21\end{aligned}

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