Question:
If the points \((1, 1, \lambda)\) & \((-3, 0, 1)\), are equidistant from the plane, \(3x + 4y - 12z + 13 = 0\), then \(\lambda\) satisfies the equation:
If the points \((1, 1, \lambda)\) & \((-3, 0, 1)\), are equidistant from the plane, \(3x + 4y - 12z + 13 = 0\), then \(\lambda\) satisfies the equation:
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Equate the two perpendicular distances to the plane and solve \(|20 - 12\lambda| = 8\).
Updated On: Nov 11, 2026
- \(3x^2 + 10x + 7 = 0\)
- \(3x^2 + 10x - 13 = 0\)
- \(3x^2 - 10x + 7 = 0\)
- \(3x^2 - 10x + 21 = 0\)
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The Correct Option is C
Solution and Explanation
Distance of \((1,1,\lambda)\): \(\frac{|3 + 4 - 12\lambda + 13|}{13} = \frac{|20 - 12\lambda|}{13}\).
Distance of \((-3,0,1)\): \(\frac{|-9 + 0 - 12 + 13|}{13} = \frac{8}{13}\).
\(|20 - 12\lambda| = 8\): \(20 - 12\lambda = 8\) gives \(\lambda = 1\); \(20 - 12\lambda = -8\) gives \(\lambda = \frac{7}{3}\).
Equation with roots \(1\) and \(\frac{7}{3}\): \((\lambda - 1)(3\lambda - 7) = 3\lambda^2 - 10\lambda + 7 = 0\).
Answer: (3) \(3x^2 - 10x + 7 = 0\)
Distance of \((-3,0,1)\): \(\frac{|-9 + 0 - 12 + 13|}{13} = \frac{8}{13}\).
\(|20 - 12\lambda| = 8\): \(20 - 12\lambda = 8\) gives \(\lambda = 1\); \(20 - 12\lambda = -8\) gives \(\lambda = \frac{7}{3}\).
Equation with roots \(1\) and \(\frac{7}{3}\): \((\lambda - 1)(3\lambda - 7) = 3\lambda^2 - 10\lambda + 7 = 0\).
Answer: (3) \(3x^2 - 10x + 7 = 0\)
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