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The equation of the line passing through...

The equation of the line passing through `(-4, 3, 1),` parallel to the plane `x +2y-z-5=0` and intersecting the line `(x+1)/-3=(y-3)/2=(z-2)/-1`

A

`(x+4)/1=(y-3)/1=(z-1)/3`

B

`(x+4)/3=(y-3)/(-1)=(z-1)/1`

C

`(x-4)/2=(y+3)/1=(z+1)/4`

D

`(x+4)/(-1)=(y-3)/1=(z-1)/1`

Text Solution

Verified by Experts

The correct Answer is:
B
Let the DR's of line be `(l,m,n)`
Then `l+2m-n=0`
Also, the line intersects the line `(x+1)/(-3)=(y-3)/2=(z-2)/(-1)`
Anypoint on this line is `[-(1+3k),2k+3,-k+2]`
It lies on `(x+4)/l=(y-3)/m=(z-1)/n`
So, there exists a k such that `(-(1+3k)+4)/l=(2k)/m=(-k+1)/n`
Hence `3/l=1/n`
`l=3n`
Thus, `m=-n`
So, Drs of the line are (3,1,-1)
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