The distance between (0,1,−1) and the point of intersection of the line x2=y−13=z+14 and the plane x+y+z=9 is
A
√29
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B
12√29
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C
49
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D
29√29
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Solution
The correct option is A√29 x2=y−13=z+14x=2t,y=1+3t,z=−1+4tx+y+z=92t+1+3t−1+4t=99t=9t=1∴(x,y,z)=(2,4,3)distance=√4+9+16=√29Hence,optionAiscorrectanswer.