jmc

We complete the square.

Factoring $3$ out of the quadratic and linear terms gives $3x^2-24x=3(x^2-8x)$.

Since $(x-4{)}^2=x^2-8x+16$, we can write $$3(x-4{)}^2=3x^2-24x+48.$$This quadratic agrees with the given $3x^2-24x+72$ in all but the constant term. We can write

$$\begin{array}{rl}3x^2-24x+72& =(3x^2-24x+48)+24\\ & =3(x-4{)}^2+24.\end{array}$$Therefore, $a=3$, $b=-4$, $c=24$, and $a+b+c=3-4+24=\boxed{23}$.