Correct Answer - `((5)/(3) hat(i) , (2)/(3) hat(j), (2)/(3) hat(k))`
Let `vec(B) =x hat(i) + yhat(j) + zhat(k)`
Given `vec(A) = hat(i) + hat(j) + hat(k) , vec(C ) = hat(j) - hat(k)`
Also given `vec(A) xx vec(B) = vec( C)`
`rArr (z-y) hat(i) - (z-x) hat(j) + (y-x) hat(k) = hat(j) - hat(k)`
`rArr z-y =0 ,x-z = 1,y - x =-1`
Also `vec(A) ". " vec(B) =3 rArr x+y +z=3`
On solving above equations we get
`x=(5)/(3) , y =z= (2)/(3)`
`vec(B) = ((5)/(3) hat(i) , (2)/(3) hat(j) , (2)/(3) hat(k))`