If the vectors $\overrightarrow{P}=\text{a}\hat{i}+\text{a}\hat{j}+3\hat{k}\text{and}\overrightarrow{Q}=\text{a}\hat{i}-2\hat{j}-\hat{k}$ are perpendicular to each other then the positive value of 'a' is

$\overline{P}.\overline{Q}=0$

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

${\text{a}}^{\text{2}}-2\text{a}-3=0$

${\text{a}}^2-3\text{a}+\text{a}-3=0$

$\text{a}\left(\text{a}-3\right)+1\left(\text{a}-3\right)=0$

$\left(\text{a}+1\right)\left(\text{a}-3\right)=0$

$\text{a}+\text{1}=\text{0}$

$\text{a}=-1$

$\text{a}-3=0$

$\text{a}=3$

$\text{a}=3\left(\text{or}\right)-1$