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Options

A. `(x-1)/(2)=(y+2)/(-1)=(z-2)/(2)`
B. `(x+(1//2))/(1)=(y-1)/(-2)=(z-(1//2))/(1)`
C. `(x)/(1)=(y)/(-2)=(z-1)/(1)`
D. `(x+1)/(1)=(y-2)/(-2)=(z=0)/(1)`

Answer

b, c, d

Explanation

`" "x+y+z-1=0`
`" "4x+y-2z+2=0`
Therefore, the line is along the vector
`" "(hati+hatj+hatk)xx(4hati+hatj-2hatk)=3hati-6hatj+3hatk`
Let `z=k`. Then `x=k-1 and y=2-2k`
Therefore, `(k-1, 2-2k, k)` is any point on the line
Hence, `(-1, 2, 0), (0, 0, 1) and (-1//2, 1, 1//2)` are the points on the line.