In an open space, there are three utility poles perpendicular to the ground with equal heights and equally spaced bases on a straight line. A person uses one-point perspective to draw these three utility poles on a canvas. A coordinate system is set up on the canvas so that the utility poles are parallel to the $y$-axis. The three base points are $A_{1}(0,0)$, $A_{2}$, $A_{3}$, all on the line $L: x + 3y = 0$; the three top points are $B_{1}(0,3)$, $B_{2}$, $B_{3}$, all on the line $M: 2x - 3y + 9 = 0$, as shown in the figure. It is known that $\overline{A_{3}B_{3}} = 2\overline{A_{1}B_{1}}$, and by one-point perspective, the intersection of lines $A_{1}B_{3}$ and $A_{3}B_{1}$ lies on line $A_{2}B_{2}$. Let $P$ be the intersection of $L$ and $M$ (this point is also called the "vanishing point").
If $\overrightarrow{PA_{1}} = k\overrightarrow{PA_{3}}$, then the value of $k$ is $\square$. (Express as a fraction in lowest terms)