The vector (s) which is/are coplanar with vectors $\hat{i}+\hat{j}+2\hat{k}$ and $\hat{i}+2\hat{j}+\hat{k}$, and perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}$ is/are

$\text{Given: } \vec{a} = \hat{i}+\hat{j}+2\hat{k}, \quad \vec{b} = \hat{i}+2\hat{j}+\hat{k}$

$\vec{c} = \hat{i}+\hat{j}+\hat{k}$

$\text{To find: vector(s) coplanar with } \vec{a},\ \vec{b} \text{ and } \perp \vec{c}$

$\vec{v} \text{ is coplanar with } \vec{a},\ \vec{b}$

$\vec{v} \perp \vec{c} \iff \vec{v}\cdot\vec{c} = 0$

$\text{Plan: } 1.\ \vec{v} = \vec{a} + \lambda\vec{b} \ \text{(coplanar with } \vec{a},\vec{b}\text{)}$

$2.\ \text{Impose } \vec{v}\cdot\vec{c} = 0 \Rightarrow \lambda$

$3.\ \text{Match } \vec{v} \text{ with the options}$

$\vec{v} = \vec{a} + \lambda\vec{b} = (1+\lambda)\hat{i} + (1+2\lambda)\hat{j} + (2+\lambda)\hat{k}$

$\vec{v}\cdot\vec{c} = (1+\lambda) + (1+2\lambda) + (2+\lambda) = 0$

$4 + 4\lambda = 0$

$\lambda = -1 \Rightarrow \vec{v} = \vec{a} - \vec{b}$

$\vec{v} = \vec{a} - \vec{b} = (0)\hat{i} + (-1)\hat{j} + (1)\hat{k}$

$\vec{v} = -\hat{j} + \hat{k}$

$\text{Required vectors form the line } \vec{w} = t(-\hat{j}+\hat{k}),\ t \in \mathbb{R}$

$\text{For every option } \vec{w}\cdot\vec{c} = 0 \Rightarrow \text{test coplanarity}$

$(1)\ \hat{j}-\hat{k} = \vec{b}-\vec{a} = -\vec{v} \ \Rightarrow \text{parallel to } \vec{v}\ \checkmark$

$(4)\ -\hat{j}+\hat{k} = \vec{a}-\vec{b} = \vec{v} \ \Rightarrow \text{parallel to } \vec{v}\ \checkmark$

$(2)\ -\hat{i}+\hat{j} \ne r(\vec{a}-\vec{b}), \quad (3)\ \hat{i}-\hat{j} \ne r(\vec{a}-\vec{b})$

$\Rightarrow (2),\ (3) \text{ are not coplanar with } \vec{a},\ \vec{b}\ \times$

$\boxed{\text{Answer: } (1)\ \hat{j}-\hat{k} \ \text{and}\ (4)\ -\hat{j}+\hat{k}}$