Let P be the plane, which contains the line of intersection of the planes, x+y+z-6=0 and…
Mathematics · JEE Main · NTA Exams — Three Dimensional Geometry
Let P be the plane, which contains the line of intersection of the planes, \(x+y+z-6=0\) and \(2 x+3 y+z+5=0\) and it is perpendicular to the xy-plane. Then the distance of the point (0,0,256) from P is equal to:
- \(205 \sqrt{5}\)
- \(11 / \sqrt{5}\)
- \(63 \sqrt{5}\)
- \(17 / \sqrt{5}\)
Answer
(B) 11 / 5
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