Assume that $P(t) = a_n t^n + \ldots + a_1 t + a_0$, where $a_n \neq 0$. Taking $x = y = z$ in the given relation, it follows
$$3 P(x) + P(3x) = 3 P(2x).$$
Looking for the coefficient of $x^n$ in this equation, we obtain the relation $3 + 3^n = 3 \cdot 2^n$, that is $3^{n-1} = 2^n - 1$. This relation holds for $n = 1$ and $n = 2$. For $n \geq 3$ we prove by induction that $3^{n-1} > 2^n - 1$.

If $n = 0$, that is $P$ is a constant polynomial $P(t) = c$, then it follows, for example from the above, that $c = 0$.

If $n = 1$, that is $P(t) = b t$, $b \neq 0$, then an easy checking shows that it satisfies the relation in the problem.

If $n = 2$, we can choose $P(t) = a t^2$, $a \neq 0$, which also satisfies the relation.

Finally, we find that all desired polynomials are
$$P(t) = a t^2 + b t,$$
where $a, b \in \mathbb{R}$.