Let $ \hat{a} $ be a unit vector perpendicular to the vectors $ \vec{b} = \hat{i} - 2\hat{j} + 3\hat{k} $ and $ \vec{c} = 2\hat{i} + 3\hat{j} - \hat{k} $, and $ \hat{a} $ makes an angle of $ \cos^{-1} \left( -\frac{1}{3} \right) $ with the vector $ \hat{i} + \hat{j} + \hat{k} $. If $ \hat{a} $ makes an angle of $ \frac{\pi}{3} $ with the vector $ \hat{i} + \alpha\hat{j} + \hat{k} $, then the value of $ a $ is:

<p>$$\begin{aligned} \text { Let } & \overrightarrow{\mathrm{v}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \begin{aligned} \overrightarrow{\mathrm{b}} & \times \overrightarrow{\mathrm{c}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{array}\right| \\ & =-7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\ & =-7(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}) \end{aligned} \end{aligned}$$</p> <p>Now $\hat{a}=\frac{\hat{\mathbf{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{3}}$ or $\frac{-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}$</p> <p>$\qquad\qquad\downarrow \qquad \qquad\downarrow$</p> <p>$$\begin{aligned} \cos \theta= & \frac{\hat{\mathrm{a}} \cdot \overrightarrow{\mathrm{v}}}{|\overrightarrow{\mathrm{v}}|}=\frac{1-1-1}{\sqrt{3} \sqrt{3}}=\frac{-1}{3} \quad \cos \theta=\frac{\hat{\mathrm{a}} \cdot \overrightarrow{\mathrm{v}}}{|\overrightarrow{\mathrm{v}}|}=\frac{-1+1+1}{3}=\frac{1}{3}\quad\text{(rejected)}\\ & \Rightarrow \hat{\mathrm{a}}=\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{3}} \end{aligned}$$</p> <p>$$\begin{aligned} & \text { Now } \cos \frac{\pi}{3}=\frac{\hat{\mathrm{a}} \cdot(\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}})}{\sqrt{1+\alpha^2+1}} \\ & \Rightarrow \frac{1}{2}=\frac{1-\alpha-1}{\sqrt{3} \sqrt{\alpha^2+2}} \\ & \Rightarrow \frac{\sqrt{3}}{2} \sqrt{\alpha^2+2}=-\alpha \quad(\therefore \alpha<0) \\ & 3 \alpha^2+6=4 \alpha^2 \\ & \Rightarrow \alpha=-\sqrt{6} \end{aligned}$$</p>