Let $\bar{a}=2 \hat{i}+\hat{j}+\hat{k}, \bar{b}=\hat{i}+2 \hat{j}-\hat{k}$ and vector $\bar{c}$ be coplanar. If $\bar{c}$ is perpendicular to $\bar{a}$, then $\overline{\mathrm{c}}$ is

Given vectors $\bar{a}=2 \hat{i}+\hat{j}+\hat{k}$ and $\bar{b}=\hat{i}+2 \hat{j}-\hat{k}$, find vector $\bar{c} = x \hat{i} + y \hat{j} + z \hat{k}$ that is coplanar with $\bar{a}$ and $\bar{b}$ and perpendicular to $\bar{a}$.

Coplanarity requires $\bar{c} \cdot (\bar{a} \times \bar{b}) = 0$. Compute $\bar{a} \times \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = \hat{i}(-1 - 2) - \hat{j}(-2 - 1) + \hat{k}(4 - 1) = -3\hat{i} + 3\hat{j} + 3\hat{k}$.

Dot product gives $-3x + 3y + 3z = 0$; dividing by 3 yields $-x + y + z = 0$.

Perpendicularity requires $\bar{c} \cdot \bar{a} = 0$, giving $2x + y + z = 0$.

Subtracting the first equation from the second: $(2x + y + z) - (-x + y + z) = 0 \implies 3x = 0 \implies x = 0$.

Substitute $x = 0$ into $-x + y + z = 0$ to get $y + z = 0$, so $z = -y$.

Thus, $\bar{c} = y(\hat{j} - \hat{k})$. Comparing with the options, $\boxed{\text{D}}$ gives $-\hat{j} + \hat{k} = -1(\hat{j} - \hat{k})$, which matches the form with $y = -1$.