Let (*) be the given functional equation.
Setting $x = y = z = 0$ in (*), we have $f(0) = 0$.
Setting $y = z = 0$ in (*) and simplifying, we have
$$f(\alpha x) = \alpha^2 f(x) \text{ for all } x \in \mathbb{R}.$$
Letting $z = 0$ in the functional equation, we get
$$f(\alpha x + y) + f(\alpha y) + f(x) = \alpha f(x + y) + 2f(x) + 2f(y).$$
Using $f(\alpha y) = \alpha^2 f(y) = (\alpha + 1)f(y)$, it follows that
$$f(\alpha x + y) = \alpha f(x + y) + f(x) + (1 - \alpha)f(y).$$
Then (*) simplifies to
$$f(x + y) + f(y + z) + f(z + x) = f(x + y + z) + f(x) + f(y) + f(z).$$
Putting $z = -y$ in the above equation, we have
$$f(x + y) + f(x - y) = 2f(x) + f(y) + f(-y).$$
Let $f_c(x) = (f(x) + f(-x))/2$ and $f_o(x) = (f(x) - f(-x))/2$. Then
$$\begin{aligned} f_c(x + y) + f_c(x - y) &= 2f_c(x) + 2f_c(y) \\ f_o(x + y) + f_o(x - y) &= 2f_o(x) \end{aligned}$$
which respectively are the quadratic and the Jensen functional equations with the continuous solutions $f_c(x) = ax^2$ and $f_o(x) = bx$ (note that $f_o(0) = 0$). Thus, $f(x) = ax^2 + bx$.
Substituting $f(x) = ax^2 + bx$ into (*), we can see that $b = 0$.
Thus $f(x) = ax^2$ is the only continuous solution.