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If the line, (x-3)/2=(y+2)/(-1)=(z+4)/3 ...

If the line, `(x-3)/2=(y+2)/(-1)=(z+4)/3` lies in the place, `l x+m y-z=9` , then `l^2+m^2` is equal to: (1) 26 (2) 18 (3) 5 (4) 2

A

26

B

18

C

5

D

2

Text Solution

Verified by Experts

The correct Answer is:
D
Since, the line `(x-3)/(2)=(y+2)/(-1)=(z+4)/(3)` line in the plane
`lx+my-z=9,` therefore we have `2l-m-3=0`
[`because` normal will be perpendicular to the line]
`implies" "2l-m=3" "...(i)`
and `" "3l-2m+4=9`
[`because` point (3, -2, -4) lies on the plane]
`implies" "3l-2m=5" "...(ii)`
On solving Eqs. (i) and (ii), we get
`l=1" and "m=-1`
`:." "l^(2)+m^(2)=2`
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