Let a plane passing through point `(-1 ,1 ,1)` is parallel to the vector `2hati+3hatj-7hatk` and the line `barr=(hati-2hatj-hatk)+lamda(3hati-8hatj+2hatk)` . The vector equation of plane is

Explanation

The line `barr = (hati - 2hatj - hatk) + lambda(3hati - 8hatj + 2hatk)`
is parallel to the vector `barb = 3hati - 8hatj + 2hatk`.
Since the plane is parallel to the vector
`bara = 2hati + 3hatj - 7hatk` and the line, the normal vector `barn` to the plane is perpendicular to both the vector `bara` and `barb`.
`thereforebarn=baraxxbarb=|{:(hati,,hatj,,hatk),(2,,3,,"-7"),(3,,-8,,2):}|`
` = (6-56)hati - (4+21)hatj + (-16-9)hatk`
` = - 50 hati - 25hatj - 25hatk`
The vector equation of the plane passing through `P(barp)` and perpendicular to `barn` is
`barr*barn = barp * barn`
Here `barp = -hati + hatj + hatk`
`thereforebarp*barn=(-hati +hatj+hatk)*(-50hati-25hatj-25hatk)`
` = (-1)(-50)+1(-25)+1(-25)`
` = 50-25-25 = 0`
`therefore` the vector equation of the required plane is
`barr* (-50 hati - 25hatj - 25hatk) = 0`
`therefore barr*(2hati + hatj + hatk) =0`.