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

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