MHT CET · Maths · Three Dimensional Geometry
If the sum of the squares of the distances of a point \(P(x, y, z)\) from the three co-ordinate axes is 324 , then the distance of point \(P\) from the origin is ....
- A \(18\)
- B \(162\)
- C \(9 \sqrt{2}\)
- D \(324\)
Answer & Solution
Correct Answer
(C) \(9 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\((y^2+z^2) + (x^2+z^2) + (x^2+y^2) = 324\) \(2(x^2+y^2+z^2) = 324\)
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