Let $\vec a = 3\hat i+\hat j-4\hat k$ and $\vec b= 6\hat i+5\hat j-2\hat k$ be two vectors. Then a vector perpendicular to a and with magnitude 3 units is

$2\hat i-2\hat j+\hat k$

The correct answer is Option (2) → $2\hat i-2\hat j+\hat k$

Given $\vec{a} = 3\hat{i} + \hat{j} - 4\hat{k}$ and $\vec{b} = 6\hat{i} + 5\hat{j} - 2\hat{k}$

A vector perpendicular to $\vec{a}$ can be taken along $(\vec{a} \times \vec{b})$.

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -4 \\ 6 & 5 & -2 \end{vmatrix} = \hat{i}(1\cdot(-2)-(-4)\cdot5) - \hat{j}(3\cdot(-2)-(-4)\cdot6) + \hat{k}(3\cdot5-1\cdot6)$

$= \hat{i}(-2+20) - \hat{j}(-6+24) + \hat{k}(15-6)$

$= 18\hat{i} - 18\hat{j} + 9\hat{k}$

$|\vec{a} \times \vec{b}| = \sqrt{18^{2}+(-18)^{2}+9^{2}} = \sqrt{324+324+81} = \sqrt{729} = 27$

Unit vector perpendicular to $\vec{a}$ (and $\vec{b}$): $\hat{n} = \frac{1}{27}(18\hat{i} - 18\hat{j} + 9\hat{k}) = \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{1}{3}\hat{k}$

Required vector with magnitude 3 units:

$\vec{r} = 3\hat{n} = 3\left(\frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{1}{3}\hat{k}\right)$

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

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