Correct Answer - `d=7(hat(i)-hat(j)-hat(k))`
Let `vec(d)=d_(1) hat(i)+d_(2) hat(j)+d_(3)hat(k)`. Then,
`vec(b).vec(a)=0, vec(d).vec(b)=0` and `vec(d).vec(c)=21`
`implies {(4d_(1)+5d_(2)-d_(3)=0),(d_(1)-4d_(2)+5d_(3)=0):}}` and `3d_(1)+d_(2)-d_(3)=21`
`implies d_(1)/((25-4))=d_(2)/((-1-20))=d_(3)/((-16-5))=k` (say) `3d_(1)+d_(2)-d_(3)=21`
`implies (d_(1)=21k, d_(2)=-21k, d_(3)=-21k)` and `3d_(1)+d_(2)-d_(3)=21`
`implies 63 k-21 k+21k=21 implies k=1/3 implies d_(1)=7, d_(2)=-7` and `d_(3)=-7`.