Are there a finite group $G$ and a field $\mathbb{F}$ such that $\gcd(3,|G|)=1$ and the group algebra $\mathbb{F}[G]$ contains a zero divisor whose support is of size $3$?

Recall that the support of an element $\alpha$ of $\mathbb{F}[G]$ is the set $\{x\in G \;|\; \alpha(x)\neq 0\}$.

This question is related to the following one:

Zero divisors of the form $1+x+y$ in the rational group algebra

One motivation to propose the question is the following well-known observation: if $a$ is an element of order $3$ in a group $G$ then $1+a+a^2$ is a zero divisor over any group algebra of $G$ (whose support is of size $3$). So the hypothesis $\gcd(3,|G|)=1$.

Are there a finite group $G$ and a field $\mathbb{F}$ such that $\gcd(3,|G|)=1$ and the group algebra $\mathbb{F}[G]$ contains a zero divisor whose support is of size $3$?

Recall that the support of an element $\alpha$ of $\mathbb{F}[G]$ is the set $\{x\in G \;|\; \alpha(x)\neq 0\}$.

This question is related to the following one:

Zero divisors of the form $1+x+y$ in the rational group algebra

One motivation to propose the question is the following well-known observation: if $a$ is an element of order $3$ in a group $G$ then $1+a+a^2$ is a zero divisor over any group algebra of $G$ (whose support is of size $3$). So the hypothesis $\gcd(3,|G|)=1$.