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Question

If foot of the perpendicular of P(2,-3,1) on the line
x+12=y33=z+21is
Q(a,b,c) then find the value of 14(a+b+c)


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Solution

Given line is x+12=y33=z+21=r...(1)
Point P(2,3,1)
Co-ordinates of foot of the perpendicular on line (1) may be taken as Q(2r1,3r+3,r2)
We get direction ratios of PQ=2r3,3r+6,r3
Direction ratios of line segment are 2,3,1 from (1)
Since PQ perpendicular to AB
2(2r3)+3(3r+6)1(r3)=0
or,14r+15=0
r=1514
Q(2r1,3r+3,r2)=(227,314,1314)
=(a,b,c) given
14(a+b+c)=14(227+1314+1314)=44+3+13=60


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