The answer is $6$. Indeed, assuming $3\nmid c-a$ and $2\nmid c-b$ we must have $4-1|(c-a)M\implies 3|M$ and $4-2|(c-b)M\implies 2|M$ and so $6|M$. Now, by Lagrange interpolation, we take the parabola that satisfies the three equations. In the denominators of the formula there will be $(4-1)$, $(4-2)$, $(2-1)$ which all get cancelled by $M$, since it is a multiple of $6$, and so the polynomial has integer coefficients. Therefore $M=6$ is the minimal such number

Note that we must have $2\mid M(c-b)$ and $3\mid M(c-a)$, thus $6\mid M$, taking $c=3,b=2,a=1$. Now, we claim that $M=6$ works. Indeed, consider polynomial $$P_{abc}=(2a+c-3b)x^2+(15b-12a-3c)x+(16a+2c-12b).$$ We are done.

GYATTERRIFIC PROBLEM!!! Using polynomial and divisibility, it is clear that 3 | P(4)-P(1) = M(c-a) and 2 | P(4)-P(2) = M(c-b). It is clearly possible to make 3 not a factor of c-a, and 2 a factor of c-b, by making c and a different values mod 3 and c and b different values mod 2. Then, it is necessary that 2|M and 3|M so 6|M. Now M=6 works: Either the solution is a linear function which has clearly integer coefficients, or it is quadratic. Consider dx^2+ex+f a quadratic equation, so that d+e+f=6a, 4d+2e+f=6b, 9d+3e+f=6c Solving this system of simultaneous equations gives the coefficients d, e, f as integers. Then M=6 is possible too.