The point of intersection of lines
$L_1=x-y-7=0$ and $L_2 \equiv x+y-5=0$ is $(6,-1)$ Now, for point $A\left(x_1, y_1\right)$ such that $P A=3 \sqrt{2}$, take the equation of line $L_1$ in symmetric form
$L_1: \frac{x-6}{\cos \frac{\pi}{4}}=\frac{y+1}{\sin \frac{\pi}{4}}= \pm 3 \sqrt{2}$
$\Rightarrow \quad x_1=6 \pm 3$ and $y_1=-1 \pm 3$
$\because \quad x_1, y_1 \geq 0$, so $A\left(x_1, y_1\right)$ is $(9,2)$
Similarly, for point $B\left(x_2, y_2\right)$ such that $P B=\sqrt{2}$ take the equation of line $L_2$ in symmetric form
$L_2: \frac{x-6}{-\cos \frac{\pi}{4}}=\frac{y+1}{\sin \frac{\pi}{4}}= \pm \sqrt{2}$
$\Rightarrow \quad x_2=6 \mp 1$ and $y_2=-1 \pm 1$
$\because x_2, y_2 \geq 0$, so $B\left(x_2, y_2\right)$ is $(5,0)$
$\therefore$ Slope of line joining origin and point $A$ is $m_1=\frac{2}{9}$ and slope of line joining origin and point $B$ is $m_2=0$
$\therefore \angle B O A=\theta=\tan ^{-1}\left(\frac{2 / 9-0}{1+0}\right)=\tan ^{-1}\left(\frac{2}{9}\right)$
$=\cos ^{-1}\left(\frac{9}{\sqrt{85}}\right)$