JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the equation of the plane passing through the line of intersection of the planes \(2 x-y+z=3,4 x-3 y\) \(+5 z+9=0\) and parallel to the line \(\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}\) is \(a x+b y+c z+6=0\). then a \(+ b + c\) is equal to \(.............\).
- A \(14\)
- B \(12\)
- C \(13\)
- D \(15\)
Answer & Solution
Correct Answer
(A) \(14\)
Step-by-step Solution
Detailed explanation
Equation of family of plane \((2 x-y+z-3)+\lambda(4 x-3 y+5 z+9)=0\) \(x(2+4 \lambda)-y(1+3 \lambda)+z(1+5 \lambda)-3+9 \lambda=0\) Parallel to the line \(-2(2+4 \lambda)-(1+3 \lambda) 4+(1+5 \lambda) 5=0\) \(5 \lambda=3\) \(\lambda=\frac{3}{5}\) equation of plane…
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