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Find the point of intersection of line passing through `(0,0,1)` and the intersection lines `x+2u+z=1,-x+y-2za n dx+y=2,x+z=2` with the `x y` plane.

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Equation of line is
`" "x+2y+z-1+lamda(-x+y-2z-2)=0`
and `" "x+y-2+mu(x+z-2)=0`
This line passes through the point (0, 0, 1). Thus,
`" "lamda=0, mu=-2`
Hence, equation of line is `x+2y+z-1=0=-x+y-2z+2`
For its point of intersection with `xy` plane, put `z=0`.
`therefore " "x+2y=1, -x+y+2=0`
Solving we get point of intersection as `((5)/(3),(-1)/(3),0)`
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