$\vec{a}$ = $3\hat{i} + 2\hat{j} + 2\hat{k}$ , $\vec{b}$ = $\hat{i} + 2\hat{j} - 2\hat{k}$ ∴ $\vec{a} + \vec{b}$ = $4\hat{i} + 4\hat{j}$ and $\vec{a} - \vec{b}$ = $2\hat{i} + 4\hat{k}$ $(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})$ = $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix}$ = $\hat{i}$ (16) - $\hat{j}$ (16) + $\hat{k}$ (-8) = $16\hat{i} - 16\hat{j} - 8\hat{k}$ ∴ $\left| (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) \right|$ = $\sqrt{16^2 + (-16)^2 + (-8)^2}$ = $\sqrt{256 + 256 + 64}$ = $\sqrt{576}$ = 24 So the unit vector, perpendicular to each of the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ is given by ± $\frac{(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})}{\left| (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) \right|}$ = ± $\frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24}$ = ± $\frac{2\hat{i} - 2\hat{j} - \hat{k}}{3}$ = ± $\frac{2}{3\hat{i}} \mp \frac{2}{3\hat{j}} \mp \frac{1}{3\hat{k}}$