Distance between two non-intersecting planes P 1 and P 2 is 5 units, where P 1 is 2x - 3y + 6z + 26 = 0 and P 2 is 4x + by + cz + d = 0. The point A (-3, 0, -1) lies between the planes P 1 and P 2 , then the value of 3b + 4c - 5d is equal to
Text Solution
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120
Since both the planes are parallel
P 1 : 4x - 6y + 12z + 52 = 0
P 2 : 4x - 6y + 12z + d = 0
Therefore, 
|d-52| = 70
d = 122,-18
P 2 is 4x - 6y + 12z + 122 = 0 or
4x - 6y + 12z - 18 = 0
Since the point (-3,0,-1) is lying between P 1 and P 2
On substituting the point in both the equations of the plane, both the expressions must be of opposite signs
Hence, 4x - 6y + 12z -18 = 0 is the equation of the required plane
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