Given,

$\overrightarrow{p}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+\hat{\mathrm{k}}$ 

$\overrightarrow{\mathit{q}}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}}$

So, $\overrightarrow{p}+\overrightarrow{q}=3\hat{\mathrm{i}}+5\hat{\mathrm{j}}+2\hat{\mathrm{k}} \& \overrightarrow{p}-\overrightarrow{q}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$

Now, $\left(\overrightarrow{p}+\overrightarrow{q}\right)\times \left(\overrightarrow{p}-\overrightarrow{q}\right)=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 3& 5& 2\\ 1& 1& 0\end{array}\right|$

$=\hat{\mathrm{i}}\left(0-2\right)-\hat{\mathrm{j}}\left(0-2\right)+\hat{\mathrm{k}}\left(3-5\right)$

$=-2\hat{\mathrm{i}}+2\hat{\mathrm{j}}-2\hat{\mathrm{k}}$

$\Rightarrow \overrightarrow{r}=\pm \sqrt{3}\frac{\left(\right(\overrightarrow{p}+\overrightarrow{q})\times (\overrightarrow{p}-\overrightarrow{q}\left)\right)}{\left|\right(\overrightarrow{p}+\overrightarrow{q})\times (\overrightarrow{p}-\overrightarrow{q}\left)\right|}=\pm \frac{\sqrt{3}(-2\hat{\mathrm{i}}+2\hat{\mathrm{j}}-2\hat{\mathrm{k}})}{\sqrt{2^2+2^2+2^2}}$

$\overrightarrow{r}=\pm \left(-\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)$

According to question 

$\overrightarrow{\mathit{r}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}$

So $\left|\alpha \right|=1,\left|\beta \right|=1,\left|\gamma \right|=1$

$\Rightarrow \left|\alpha \right|+\left|\beta \right|+\left|\gamma \right|=3$