The sum of all the values of \(\lambda\) for which the distance of the point P (2, 3, \(\lambda\)) from the plane x + 2y − 2z = 9 is 3 units, is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → −1
The distance of a point from a plane ax + by + cz = d is
\( \displaystyle \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}} \)
For P(2, 3, \(\lambda\)):
\( \displaystyle \frac{|2 + 6 - 2\lambda - 9|}{\sqrt{1 + 4 + 4}} = 3 \)
\(|-1 - 2\lambda| = 9\)
\(2\lambda + 1 = 9 \text{ or } 2\lambda + 1 = -9\)
\(\lambda = 4 \text{ or } \lambda = -5\)
Sum of the values:
\(4 + (-5) = -1\)
Note: This topic was not in the CUET 2026 syllabus.