Letting $a = 0$ and $b = c = 2$ we obtain $2 \ge M$. We will show that $M = 2$ works.

Without loss of generality we may assume that $\max\{a, b, c\} = c$. Let $x = a + b$ and $y = ab$. We have $c \ge \frac{a+b+c}{3} = \frac{4}{3}$ and hence $x = a+b \le \frac{8}{3}$.

Then
$$a^2 + b^2 + c^2 + 3abc \ge 2(ab + bc + ca) \iff x^2 - 2y + (4-x)^2 + 3y(4-x) \ge 2(y+x(4-x)) \iff 4(x-2)^2 + y(8-3x) \ge 0$$
follows.