Distance of point (1,2,3) from plane x+y+z−6=0 is:
Distance of point (1,2,3) from plane x+y+z−6=0 is:
- A. 0
- B. 1/√3
- C. 2/√3
- D. √3
Answer: A) 0
Explanation: Substituting gives 1+2+3−6=0. Point lies on plane. Distance=0.
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