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Question

The distance between the point (1,5,10) and the point of intersection of the line x23=y+14=z212 with the plane xy+z=5 is

A
13
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B
1
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C
2
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D
13
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Solution

The correct option is A 13
Given line is x23=y+14=z212
Any point on the line is
(3λ+2,4λ1,12λ+2)
On substituting in the equation of the plane, we get
3λ+2(4λ1)+(12λ+2)=5λ=0
Point of intersection =(2,1,2)
The required distance
=(2+1)2+(1+5)2+(2+10)2=13

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