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

Let the equation of the plane, that passes through the point \((1,4,-3)\) and contains the line of intersection of the planes \(3 x-2 y+4 z-7=0\) and \(x+5 y-2 z+9=0\), be \(\alpha x+\beta y+\gamma z+3=0\), then \(\alpha+\beta+\gamma\) is equal to :

  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 \((1,4,-3)\) \(\Rightarrow 3+\lambda+20 \lambda-8-12+6 \lambda+9 \lambda-7=0\) \(\Rightarrow \lambda=\frac{2}{3}\)…