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The correct Answer is:
`9//2` cubic units

Since the plane is parallel to lines `L_(1) and L_(2)` with direction ratios as `(1, 0, -1) and (1, -1, 0)`, a vector perpendicular to `L_(1) and L_(2)` will be parallel to the nomal `vecn` to the plane. Therefore,
`" "vecn= |{:(hati,,hatj,,hatk),(1,,0,,-1),(1,,-1,,0):}|=-hati-hatj-hatk`
The equation of the plane passing through `(1, 1, 1)` and having normal vector `vecn=-hati-hatj-hatk` is given by
`" "(vecr-veca)*vecn=0`
`rArr" "-1(x-1)-1(y-1)-1(z-1)=0`
`" "x+y+z=3`
`" "(x)/(3)+ (y)/(3)+(z)/(3)=1`
The plane meets the axes at `A(3, 0, 0), B(0, 3, 0) and C(0, 0, 3) or A(3hati), B(3hatj) and C(3hatk)`.
Hence, the volume of tetrahedron
`" "OABC= (1)/(6) [ 3hati3hatj3hatk]`
`" "=(27)/(6)= (9)/(2)` cubic units
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