4. In the Cartesian coordinate system $x O y$, line $l$ passes through the origin, and $\vec{n}=(3,1)$ is a normal vector of $l$. It is known that the sequence $\left\{a_{n}\right\}$ satisfies: for any positive integer $n$, the point $\left(a_{n+1}, a_{n}\right)$ lies on $l$. If $a_{2}=6$, then the value of $a_{1} a_{2} a_{3} a_{4} a_{5}$ is $\qquad$ .
According to the problem, the line $l: 3 x+y=0 \Rightarrow 3 a_{n+1}+a_{n}=0 \Rightarrow a_{n+1}=-\frac{1}{3} a_{n}$, thus the sequence $\left\{a_{n}\right\}$ is a geometric sequence with a common ratio of $-\frac{1}{3}$. Therefore, $a_{1} a_{2} a_{3} a_{4} a_{5}=a_{3}^{5}=(-2)^{5}=-32$.