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Question

Let the plane ax+by+cz+d=0 bisect the line joining the points (4,3,1) and (2,3,5) at the right angles. If a,b,c,d are integers, then the minimum value of (a2+b2+c2+d2) is

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Solution

Normal of plane =PQ=2^i+6^j6^k
a=2,b=6,c=6
Equation of plane is
2x+6y6z+d=0


Mid point of P(4,3,1) and Q(2,3,5) is M(3,0,2).
Since, plane passes through M,
d=6
So, equation of plane is
2x+6y6z6=0
x3y+3z+3=0
(a2+b2+c2+d2)min=12+9+9+9=28

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