JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the points \((1, 1, \lambda )\) and \((-3, 0, 1)\) are equidistant from the plane, \(3x + 4y - 12z + 13 = 0,\) then \(\lambda \) satisfies the equation
- A \(3x^2 +10x - 13 =0\)
- B \(3x^2 -10x+21=0\)
- C \(3x^2 - 10x+7 =0\)
- D \(3x^2+10x - 7=0\)
Answer & Solution
Correct Answer
(C) \(3x^2 - 10x+7 =0\)
Step-by-step Solution
Detailed explanation
\(|3+4-12 \lambda+13|=|-9+0-12+13|\) \(\Rightarrow|-12 \lambda+20|=|8|\) \(\Rightarrow|3 \lambda-5|=2\) \(\Rightarrow 9 \lambda^{2}+25-30 \lambda=4\) \(\Rightarrow 9 \lambda^{2}-30 \lambda+21=0\) \(\Rightarrow 3 \lambda^{2}-10 \lambda+7=0\)
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