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Question

The distance of the point (1, 0, 2) from the point of intersection of the line
x23=y+14=z212 and the plane x - y + z = 16 is


A

214

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B

8

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C

321

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D

13

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Solution

The correct option is D

13


Given equation of line is
x23=y+14=z212=λ [say]...(i)
and equation of plane is
xy+z=16 .........(ii)
Any point on the line (i) is, (3λ+2,4λ1,12λ+2)
Let this point of intersection of the line and plane.
(3λ+2)(4λ1)+(12λ+2)=16 11λ+5=16 11λ=11λ=1
So, the point of intersection is (5, 3, 14).
Now, distance between the points (1, 0, 2) and (5, 3, 14)
=(51)2+(30)2+(142)2
=16+9+144=169=13


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