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JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(\lambda_1, \lambda_2\) be the values of \(\lambda\) for which the points \(\left(\frac{5}{2}, 1, \lambda\right)\) and \((-2,0,1)\) are at equal distance from the plane \(2 x+3 y-6 z+7=0\). if \(\lambda_1 > \lambda_2\), then the distance of the point \(\left(\lambda_1-\lambda_2, \lambda_2, \lambda_1\right)\) from the line \(\frac{x-5}{1}=\frac{y-1}{2}=\frac{z+7}{2}\) is \(............\).
- A \(10\)
- B \(9\)
- C \(12\)
- D \(13\)
Answer & Solution
Correct Answer
(B) \(9\)
Step-by-step Solution
Detailed explanation
\(2 x+3 y-6 z+7=0\left(\frac{5}{2}, 1, \lambda\right),(-2,0,1)\) \(d _1=\left|\frac{5+3-6 \lambda+7}{7}\right|= d _2=\left|\frac{-4-6+7}{7}\right|\) \(\Rightarrow|15-6 \lambda|=|3|\) \(15-6 \lambda=3 \text { or } 15-6 \lambda=-3\) \(6 \lambda=12\) \(\lambda=2\) \(\lambda=3\)…
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