JEE Math Practice Question
Let P be the plane, which contains the line of intersection of the planes, x + y + z – 6 = 0 and 2x + 3y + 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 :
- A.205$$\sqrt5$$
- B.63$$\sqrt5$$
- C.11/$$\sqrt5$$
- D.17/$$\sqrt5$$
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