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

माना समतल, जो बिन्दु \((1,4,-3)\) से होकर जाता है तथा जिसमें समतलों \(3 x -2 y +4 z -7=0\) तथा \(x +5 y -2 z +9=0\), की प्रतिच्छेदन रेखा स्थित है, का समीकरण \(\alpha x +\beta y +\gamma z +3=0\), है, तो \(\alpha+\beta+\gamma\) बराबर है 

  1. A \(-23\)
  2. B \(-15\)
  3. C \(23\)
  4. D \(15\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(-23\)

Step-by-step Solution

Detailed explanation

Equation of plane is \(3 x-2 y+4 z-7+\lambda(x+5 y-2 z+9)=0\) \((3+\lambda) x+(5 \lambda-2) y+(4-2 \lambda) z+9 \lambda-7=0\) passing through \((1,4,-3)\) \(\Rightarrow 3+\lambda+20 \lambda-8-12+6 \lambda+9 \lambda-7=0\) \(\Rightarrow \lambda=\frac{2}{3}\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app