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

Given,

$$\begin{gathered} \overrightarrow{P}=a\hat{i}+a\hat{j}+3\hat{k} \\ \overrightarrow{Q}=a\hat{i}-2\hat{j}-\hat{k} \end{gathered}$$

When two vectors are parallel to one another, then

$$\begin{gathered} \overrightarrow{P}.\overrightarrow{Q}=0 \\ (a\hat{i}+a\hat{j}+3\hat{k}).(a\hat{i}-2\hat{j}-\hat{k})=0 \\ \Rightarrow a^2-2a-3=0 \end{gathered}$$

By solving the above quadratic equation,

we get, a=3 or a=-1

The positive value is 3.

Hence the correct answer is 3.