Given that $\overrightarrow{a}·\overrightarrow{b}=\overrightarrow{a}·\overrightarrow{c}=0$

$\Rightarrow$$\overrightarrow{a}$ is perpendicular to both $\overrightarrow{b} \& \overrightarrow{c}$.

The vector perpendicular to both $\overrightarrow{b} \& \overrightarrow{c}$ is $\left(\overrightarrow{b}\times \overrightarrow{c}\right)$.

Now, unit vector $\overrightarrow{a}=\frac{\left(\overrightarrow{b}\times \overrightarrow{c}\right)}{\left|\overrightarrow{b}\times \overrightarrow{c}\right|}$.

We know that, ${\left|\overrightarrow{b}\times \overrightarrow{c}\right|}^2={\left|\overrightarrow{b}\right|}^2{\left|\overrightarrow{c}\right|}^2-{\left|\overrightarrow{b}·\overrightarrow{c}\right|}^2$

$\Rightarrow {\left|\overrightarrow{b}\times \overrightarrow{c}\right|}^2=1·1-{\left(\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|\cos \frac{\pi }{3}\right)}^2$

$\Rightarrow {\left|\overrightarrow{b}\times \overrightarrow{c}\right|}^2=1-{\left(\frac{1}{2}\right)}^2=\frac{3}{4}$

Then, $\overrightarrow{a}=\pm \frac{2}{\sqrt{3}}\left(\overrightarrow{b}\times \overrightarrow{c}\right)$

Now, $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}{]}^2={\left(\overrightarrow{a}.\left(\overrightarrow{b}\times \overrightarrow{c}\right)\right)}^2$

$={\left[\pm \frac{2}{\sqrt{3}}\left(\overrightarrow{b}\times \overrightarrow{c}\right)·\left(\overrightarrow{b}\times \overrightarrow{c}\right)\right]}^2$

$={\left[\pm \frac{2}{\sqrt{3}}{\left|\overrightarrow{b}\times \overrightarrow{c}\right|}^2\right]}^2$

$={\left[\pm \frac{2}{\sqrt{3}}\times \frac{3}{4}\right]}^2$

$=\frac{3}{4}$