Let $\overrightarrow p = 2\widehat i + 3\widehat j + \widehat k$ and $\overrightarrow q = \widehat i + 2\widehat j + \widehat k$ be two vectors. If a vector $\overrightarrow r = (\alpha \widehat i + \beta \widehat j + \gamma \widehat k)$ is perpendicular to each of the vectors ($(\overrightarrow p + \overrightarrow q )$ and $(\overrightarrow p - \overrightarrow q )$, and $\left| {\overrightarrow r } \right| = \sqrt 3$, then $\left| \alpha \right| + \left| \beta \right| + \left| \gamma \right|$ is equal to _______________.

$\overrightarrow p = 2\widehat i + 3\widehat j + \widehat k$ (Given )<br><br>$\overrightarrow q = \widehat i + 2\widehat j + \widehat k$<br><br>Now, $$(\overrightarrow p + \overrightarrow q ) \times (\overrightarrow p - \overrightarrow q ) = \left| {\matrix{ {\widehat i} &amp; {\widehat j} &amp; {\widehat k} \cr 3 &amp; 5 &amp; 2 \cr 1 &amp; 1 &amp; 0 \cr } } \right|$$<br><br>$= - 2\widehat i - 2\widehat j - 2\widehat k$<br><br>$$ \Rightarrow \overrightarrow r = \pm \sqrt 3 {{\left( {(\overrightarrow p + \overrightarrow q ) \times (\overrightarrow p - \overrightarrow q )} \right)} \over {\left| {(\overrightarrow p + \overrightarrow q ) \times (\overrightarrow p - \overrightarrow q )} \right|}} = \pm {{\sqrt 3 \left( { - 2\widehat i - 2\widehat j - 2\widehat k} \right)} \over {\sqrt {{2^2} + {2^2} + {2^2}} }}$$<br><br>$$\overrightarrow r = \pm \left( { - \widehat i - \widehat j - \widehat k} \right)$$<br><br>According to question <br><br>$\overrightarrow r = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$<br><br>So, |$\alpha$| = 1, |$\beta$| = 1, |$\gamma$| = 1<br><br>$\Rightarrow$ $\left| \alpha \right| + \left| \beta \right| + \left| \gamma \right|$ = 3