1.

Thedistance of the point (1, 0, 2) from the point of intersection of the line `(x-2)/3=(y+1)/4=(z-2)/(12)`and the plane x y + z = 16, is :(1) `2sqrt(14)`(2) 8 (3) `3sqrt(21)`(4) 27A. `2sqrt(14)`B. 8C. `3sqrt(21)`D. 13

Answer» Correct Answer - D
Given equation of line is
`(x-2)/(3)=(y+1)/(4)=(z-2)/(12)=lambda" "["say"]…(i)`
and equation of plane is
`x-y+z=16" "…(ii)`
Any point on the line (i) is, `(3lambda+2,4lambda-1,12lambda+2)`
Let this point of intersection of the line and plane.
`:.(3lambda+2)-(4lambda-1)+(12lambda+2)=16`
`implies" "11lambda+5=16`
`implies" "11lambda=11`
`implies" "lambda=1`
So, the point of intersection is (5, 3, 14)
Now, distance between the points (1, 0, 2) and (5, 3, 14)
`sqrt((5-1)^(2)+(3-0)^(2)+(14-2)^(2))`
`=sqrt(16+9+144)=sqrt(169)=13`


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