JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the equation of the plane passing through the line of intersection of the planes \(2 x-7 y+4 z-3=0,3 x-5 y+4 z+11=0\) and the point \((-2,1,3)\) is \(a x+b y+c z-7=0,\) then the value of \(2 a+b+c-7\) is
- A \(9\)
- B \(12\)
- C \(4\)
- D \(8\)
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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