$\frac{1}{21}(17 \mathbf{a}-12 \mathbf{b})$

Since $O P=2 P A$, $\overrightarrow{O P}=\frac{2}{3} \mathbf{a}$
Using Ratio Theorem,
$$
\begin{array}{l}
\overrightarrow{O Q}=\frac{5 \mathbf{a}+4 \mathbf{b}}{9}=\frac{5}{9} \mathbf{a}+\frac{4}{9} \mathbf{b} \\
\overrightarrow{P Q}=\overrightarrow{O Q}-\overrightarrow{O P}=-\frac{1}{9} \mathbf{a}+\frac{4}{9} \mathbf{b}=\frac{1}{9}(4 \mathbf{b}-\mathbf{a})
\end{array}
$$
Since the line passes through $P$ and is // to $4 \mathbf{b}-\mathbf{a}$ therefore an equation of $l$ is
$$
\mathbf{r}=\frac{2}{3} \mathbf{a}+\lambda(4 \mathbf{b}-\mathbf{a}) \text {. where } \lambda \in \mathbb{R}
$$

Since $X$ lies on $l$, we have $O X=\frac{2}{3} \mathbf{a}+t(4 \mathbf{b}-\mathbf{a})$ for a particular value of $t$ Since $A X$ is perpendiellar to lo, $\overrightarrow{A X} \bullet(4 \mathbf{b}-\mathbf{a})=0$
$$
\begin{array}{l}
\left(\frac{2}{3} \mathbf{a}+t(4 \mathbf{b}-\mathbf{a})-\mathbf{a}\right) \cdot(4 \mathbf{b}-\mathbf{a})=0 \\
\left(\left(-\frac{1}{3}-t\right) \mathbf{a}+4 \mathbf{b} \mathbf{b}\right) \cdot(4 \mathbf{b}-\mathbf{a})=0
\end{array}
$$
Since $\mathbf{a}$ and $\mathbf{b}$ are perpendicular, $\therefore \mathbf{a} \cdot \mathbf{b}=0$
$$
\therefore\left(\frac{1}{3}+t\right)|\mathbf{a}|^2+16 t|\mathbf{b}|^2=0, \quad|\mathbf{a}|=\sqrt{3},|\mathbf{b}|=\frac{1}{2}
$$
$$
\therefore\left(\frac{1}{3}+t\right) 3+4 t=0
$$

$t=-\frac{1}{7}$
$\therefore \overrightarrow{O X}=\frac{2}{3} \mathbf{a}-\frac{1}{7}(4 \mathbf{b}-\mathbf{a})=\frac{1}{21}(17 \mathbf{a}-12 \mathbf{b})$