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Question

If the position vector of the point of intersection of the line r=(i+2j+3k)+λ(2i+j+2k) and the plane r(2i6j+3k)+5=0 is ai+bj+ck , then (50a+60b+75c)2 is equal to

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Solution

given eq of line
r=(^i+2^j+3^k)+λ(2^i+^j+2^k)
comparing with eq of line we get
position vector A(1,2,3) a=2,b=1,c=2
eq become
x12=y21=z32=λ
passing the given eq to pointa^i+b^j+c^k we get
a=2λ+1,b=λ+2,c=2λ+3
passing given points to eq of planes
2a6b+3c+5=0
putting values of a,b,c
2(2λ+1)6(λ+2)+3(2λ+3)+5=0
4λ+26λ12+6λ+9+5=0
4λ=4
λ=1
a=2λ+1=2+1=1
b=λ+2=1+2=1
c=2λ+3=2+3=1
(50a+60b+75c)2
(50+60+75)2
852=7225


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