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Question

The shortest distance between the plane 12x+64y+3z=327 and the sphere x2+y2+z2+4x2y6z=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 13
The 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=13

Perpendicular distance formula is
|(ax1+by1+cz1+d)(a2+b2+c2)|

= 12(2)+4(1)+3(3)327122+42+32

length of the perpendicular CP=33813=26

PQ=CPCQ=2613=13

825937_163734_ans_9cf302e75a904bfa8623c0b693422045.png

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