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Question

# The shortest distance between the plane 12x+64y+3z=327 and the sphere x2+y2+z2+4xâˆ’2yâˆ’6z=155 is

A
1134
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B
13
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C
39
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D
26
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Solution

## The correct option is B 13The shortest distance required is the perpendicular distance from the centre of the sphere to the plane.consider the above image Center of the sphere = (−2,1,3).Radius of the sphere CQ=√(−2)2+12+32−(−155) = √4+1+9+155 = √169 CQ=13Perpendicular distance formula is|(ax1+by1+cz1+d)√(a2+b2+c2)|= 12(−2)+4(1)+3(3)−327√122+42+32length of the perpendicular CP=33813=26PQ=CP−CQ=26−13=13

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