The distance between the point (−1,−5,−10) and the point of intersection of the line x−23=y+14=z−212 with the plane x−y+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 A13 Given line is x−23=y+14=z−212
Any point on the line is (3λ+2,4λ−1,12λ+2)
On substituting in the equation of the plane, we get 3λ+2−(4λ−1)+(12λ+2)=5⇒λ=0 ∴ Point of intersection =(2,−1,2) ∴ The required distance =√(2+1)2+(−1+5)2+(2+10)2=13