Given $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}, \vec{b} = 3\hat{i} - \hat{k}$ and $\vec{c} = 2\hat{i} + \hat{j} - 2\hat{k}$. Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and $\vec{c} \cdot \vec{d} = 3$.

$3\hat{i} + 15\hat{j} + 9\hat{k}$

The correct answer is Option (2) → $3\hat{i} + 15\hat{j} + 9\hat{k}$ ##

$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$

$\vec{b} = 3\hat{i} - \hat{k}$

$\vec{c} = 2\hat{i} + \hat{j} - 2\hat{k}$

$\vec{d} = (x\hat{i} + y\hat{j} + z\hat{k})$

Since, $\vec{d}$ is perpendicular to $\vec{a}$ and $\vec{b}$

$\vec{d} \cdot \vec{a} = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} - \hat{j} + \hat{k})$

$= 2x - y + z = 0$

$\vec{d} \cdot \vec{b} = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (3\hat{i} - \hat{k})$

$= 3x - z = 0$

$\vec{c} \cdot \vec{d} = (2\hat{i} + \hat{j} - 2\hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k})$

$= 2x + y - 2z = 3$

On solving $x = 3, y = 15, z = 9$

vector $\vec{d} = x\hat{i} + y\hat{j} + z\hat{k}$

$= 3\hat{i} + 15\hat{j} + 9\hat{k}$