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Question

If the shortest distance between the lines x1α=y+11=z1, (α1) and x+y+z+1=0=2xy+z+3 is 13, then a value of α is :

A
3219
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B
1932
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C
1619
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D
1916
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Solution

The correct option is A 3219
Family of plane is,
x+y+z+1+k(2xy+z+3)=0(1+2k)x+(1k)y+(1+k)z+(1+3k)=0
If given family of plane is parallel to the line then
(1+2k)α(1k)+(1+k)=0α=2k1+2kPerpendicular distance between x1α=y+11=z1 & (1+2k)x+(1k)y+(1+k)z+(1+3k)=0 is|(1+2k)(1)+(1k)(1)+(1+k)(0)+(1+3k)|(1+2k)2+(1k)2+(1+k)2=13k=0,32102α=0,3219

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