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The point on the line (x-2)/1=(y+3)/(-2)...

The point on the line `(x-2)/1=(y+3)/(-2)=(z+5)/(-2)` at a distance of 6 from the point `(2,-3,-5)` is a. `(3,-5,-3)` b. `(4,-7,-9)` c. `0,2,-1` d. none of these

A

`(3,-5,-3)`

B

`(4,-7,-9)`

C

`(0,1,-1)`

D

`(-3,5,3)`

Text Solution

Verified by Experts

The correct Answer is:
B, C

Direction cosines of the given line are `(1)/(3),-(2)/(3)-(2)/(3)` Hence, the equation of line can be a point in the form
`(x-2)/((1)/(3))=(y+3)/((-2)/(3))=(z+5)/((-2)/(3))=r`
Therefore, any point of the line is `(2+(r)/(3),-3-(2r)/(3),-5-(2r)/(3))`
where `r=+-6`
Points aer `(4,j-7,-9)` and `(0,1,-1)`
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