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

यदि समतलों \(2 x-7 y+4 z-3=0\), \(3 x -5 y +4 z +11=0\) की प्रतिच्छेदन रेखा तथा बिन्दु \((-2,1,3)\) से होकर जाने वाले समतल का समीकरण \(ax + by + cz -7=0\) है, तो \(2 a + b + c -7\) का मान है

  1. A \(9\)
  2. B \(12\)
  3. C \(4\)
  4. D \(8\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4\)

Step-by-step Solution

Detailed explanation

Required plane is \(p _{1}+\lambda p _{2}=(2+3 \lambda) x -(7+5 \lambda) y\) \(+(4+4 \lambda) z-3+11 \lambda=0\) which is satisfied by \((-2,1,3)\). Hence, \(\lambda=\frac{1}{6}\) Thus, plane is \(15 x-47 y+28 z-7=0\) So, \(2 a+b+c-7=4\)
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