JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the line \(\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}\) intersect the plane containing the lines \(\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}\) and \(4 a x-y+5 z-7 a=0=2 x -5 y - z -3, a \in R\) at the point \(P(\alpha, \beta, \gamma)\). Then the value of \(\alpha+\beta+\gamma\) equals\(...\)
- A \(13\)
- B \(11\)
- C \(12\)
- D \(10\)
Answer & Solution
Correct Answer
(C) \(12\)
Step-by-step Solution
Detailed explanation
Equation of plane \(4 a x-y+5 z-7 a+\lambda(2 x -5 y - z -3)=0\) this satisfy \((4,-1,0)\) \(16 a+1-7 a+\lambda(8+5-3)=0\) \(9 a+1+10 \lambda=0\) Normal vector of the plane A is \((4 a+2 \lambda,-1-5 \lambda, 5-\lambda)\) vector along the line which contained the plane \(A\) is…
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