Let $\vec a = \hat i + \hat j + 2\hat k = (1,1,2), \qquad \vec b = \hat i + 2\hat j + \hat k = (1,2,1).$

We need vectors which are:

• coplanar with $\vec a$ and $\vec b$ r lie in the plane spanned by $\vec a, \vec b$,
• perpendicular to $\hat i+\hat j+\hat k = (1,1,1)$.

Any vector coplanar with $\vec a$ and $\vec b$ can be written as $\vec v = x\vec a + y\vec b.$ So, $\vec v = x(1,1,2) + y(1,2,1) = (x+y,\ x+2y,\ 2x+y).$

Since $\vec v$ is perpendicular to $(1,1,1)$, $\vec v\cdot (1,1,1)=0.$ Therefore, $(x+y)+(x+2y)+(2x+y)=0$ $4x+4y=0$ $x+y=0 \implies y=-x.$

Hence $\vec v = x(\vec a-\vec b).$ Now, $\vec a-\vec b=(1,1,2)-(1,2,1)=(0,-1,1)= -\hat j+\hat k.$ So every such vector is a scalar multiple of $-\hat j+\hat k$.

Check the options:

• A: $\hat j-\hat k = -( -\hat j+\hat k)$

  This is a scalar multiple of the required vector, so it lies in the same plane and is perpendicular to $(1,1,1)$.

  Correct.

• B: $-\hat i+\hat j = (-1,1,0)$

  Dot with $(1,1,1)$: $(-1)+1+0=0,$ but check if it is in span of $\vec a,\vec b$. If $x\vec a+y\vec b=(-1,1,0),$ then $x+y=-1,\quad x+2y=1,\quad 2x+y=0.$ From first two, $y=2$, $x=-3$, then $2x+y=-6+2=-4\neq 0$. Not coplanar with $\vec a,\vec b$.

  Incorrect.

• C: $\hat i-\hat j = (1,-1,0)$

  Dot with $(1,1,1)$: $1-1+0=0,$ but check span: $x+y=1,\quad x+2y=-1,\quad 2x+y=0.$ From first two, $y=-2$, $x=3$, then $2x+y=6-2=4\neq 0$. Not coplanar.

  Incorrect.

• D: $-\hat j+\hat k = (0,-1,1)$

  This is exactly $\vec a-\vec b$, so it is coplanar. Also, $0-1+1=0,$ so it is perpendicular to $(1,1,1)$.

  Correct.

Therefore, the correct vectors are $\boxed{\hat j-\hat k \text{ and } -\hat j+\hat k}.$