TS EAMCET · Maths · Three Dimensional Geometry
Let \(6 x-3 y+2 z-6=0\) be the given plane. If \(a, b, c\) are the intercepts made by the plane on \(\mathrm{X}, \mathrm{Y}, \mathrm{Z}\)-axes respectively; \(1, m, n\) are the direction cosines of a normal drawn to the plane and \(\mathrm{p}\) is the perpendicular distance from the origin to the plane, then \(|a l+b m+c n|=\)
- A \(\mathrm{p}\)
- B \(2 \mathrm{p}\)
- C \(3 \mathrm{p}\)
- D \(4 p\)
Answer & Solution
Correct Answer
(C) \(3 \mathrm{p}\)
Step-by-step Solution
Detailed explanation
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