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Question

Find the equation of the plane which passes through the point (3, 4, −1) and is parallel to the plane 2x − 3y + 5z + 7 = 0. Also, find the distance between the two planes.

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Solution

Let the equation of a plane parallel to the given plane be 2x - 3y + 5z = k ... 1This passes through (3, 4, -1).  So, 2 3 -3 4+5 -1 = kk = -11Substituting this in (1), we get2x - 3y + 5z =-11... 1, which is the equation of the required plane.The equation of the given plane is2x - 3y + 5z = -7 ... 2We know that the distance between two planes ax+by+cz=d1 and ax+by+cz=d2 is d2-d1a2+b2+c2So, the required distance =-7 - -1122 + -32 + 52=-7 + 114 + 9 + 25= 438 units

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