$\text{ If }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are unit vectors such that }\overrightarrow{a}\cdot \overrightarrow{b}=0=\overrightarrow{a}\cdot \overrightarrow{c}\text{ and the angle between }\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ is }\frac{\pi }{3}\text{. Then the value of }|\overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{a}\times \overrightarrow{c}|\text{ is }$

$\begin{array}{l}\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}=0\Rightarrow \overrightarrow{\mathrm{a}}\perp \overrightarrow{\mathrm{b}}\\ \overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{c}}=0\Rightarrow \overrightarrow{\mathrm{a}}\perp \overrightarrow{\mathrm{c}}\\ \Rightarrow \overrightarrow{\mathrm{a}}\perp \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}\\ \begin{array}{rl}\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}}\mid & =|\overrightarrow{\mathrm{a}}\times (\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}\left)\right|\\ & =\left|\overrightarrow{\mathrm{a}}\right||\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}|=|\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}|\end{array}\end{array}$

$\begin{array}{c}\text{ Now }|\overrightarrow{b}-\overrightarrow{c}{|}^2=|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2-2⌊\overrightarrow{b}\left|\right|\overrightarrow{c}\mid \cos \frac{\pi }{3}\\ =2-2\times \frac{1}{2}=1\\ |\overrightarrow{b}-\overrightarrow{c}|=1\end{array}$