Let $\vec a =\hat i +\hat j+\hat k$ and $\vec b = \hat i+2\hat j+3\hat k$, then a unit vector perpendicular to both vectors $(\vec a +\vec b)$ and $(\vec a -\vec b)$ is equal to

$\frac{1}{\sqrt{6}}(-\hat i+2\hat j-\hat k)$

The correct answer is Option (1) → $\frac{1}{\sqrt{6}}(-\hat i+2\hat j-\hat k)$

Given:

$\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$

$\vec{a} + \vec{b} = (1+1)\hat{i} + (1+2)\hat{j} + (1+3)\hat{k} = 2\hat{i} + 3\hat{j} + 4\hat{k}$

$\vec{a} - \vec{b} = (1-1)\hat{i} + (1-2)\hat{j} + (1-3)\hat{k} = 0\hat{i} - \hat{j} - 2\hat{k}$

Vector perpendicular to both is their cross product:

$\vec{r} = (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})$

$\vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix} = \hat{i}(3\times(-2) - 4\times(-1)) - \hat{j}(2\times(-2) - 4\times0) + \hat{k}(2\times(-1) - 3\times0)$

$\vec{r} = \hat{i}(-6+4) - \hat{j}(-4) + \hat{k}(-2)$

$\vec{r} = -2\hat{i} + 4\hat{j} - 2\hat{k}$

Magnitude: $|\vec{r}| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}$

Unit vector: $\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}}$

Required unit vector = $\frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}}$