1.

On which of the following lines lies the point of intersection of the line, `(x-4)/(2)=(y-5)/(2)=(z-3)/(1)` and the plane, `x+y+z=2`?A. `(x-4)/(1)=(y-5)/(1)=(z-5)/(-1)`B. `(x+3)/(3)=(4-y)/(3)=(z+1)/(-2)`C. `(x-2)/(2)=(y-3)/(2)=(z+3)/(3)`D. `(x-1)/(1)=(y-3)/(2)=(z+4)/(-5)`

Answer» Correct Answer - D
Given equation of line is
`(x-4)/(2)=(y-5)/(2)=(z-3)/(1)=r ("let")" "…(i)`
`impliesx=2r+4,y=2r+5" and "z=r+3`
`:."General point on the line (i) is "`
`P(2r+4,2r+5,r+3)`
So, the point of intersection of line (i) and plane `x+y+z=2` will be of the form `P(2r+4,2r+5,r+3)` for some `rinR.`
`implies(2r+4)+(2r+5)+(r+3)=2`
[`because` the point will lie on the plane]
`implies5r=-10impliesr=-2`
So, the point of intersection is P(0, 1, 1)
[putting `r=-2` in `(2r+4,2r+5,r+3)]`
Now, on checking the options, we get
`(x-1)/(1)=(y-3)/(2)=(z+4)/(-5)` contain the point (0,1,1)


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