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JEE Mains · Maths · STD 12 - 11. three dimension geometry

मान लीजिए कि रेखाओं \(x+2=y-1=z, \frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}\) और \(\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}\) द्वारा बने त्रिभुज का क्षेत्रफल \(A\) है। तब \(A^2\) = ___

  1. A 55
  2. B 56
  3. C 57
  4. D 58
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Correct Answer

(B) 56

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Detailed explanation

\begin{aligned} & \mathrm{L}_1: \mathrm{x}+2=\mathrm{y}-1=\mathrm{z}=\ell \\ & \mathrm{L}_2: \frac{\mathrm{x}-3}{5}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-1}{1}=\mathrm{m} \\ & \mathrm{~L}_3: \frac{\mathrm{x}}{-3}=\frac{\mathrm{y}-3}{5}=\frac{\mathrm{z}-2}{1}=\mathrm{n}…

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