I can't test them all!

Let $f(x)=| x-3 |+| 2x+4|+| x |$ $f(x)=\begin{cases} -4x-1&x<-2\\7 &-2\leq x<0 \\2x+7& 0\leq x<3 \\4x+1& x\geq 3\end{cases}$ $f(x)\leq 11 \implies$ $f(x)=\begin{cases} -4x-1\leq 11&x<-2\\7\leq 11 &-2\leq x<0 \\2x+7\leq 11& 0\leq x<3 \\4x+1\leq 11 & x\geq 3\end{cases}$ Fourth case gives null set while from other three cases we get $-3\leq x<-2$ , $-2\leq x<0$ and $0\leq x\leq2$ respectively in which integral solutions are $-3,-2,-1,0,1,2$ . Hence a total of $\Large \boxed{6}$ .

For integer $x$ . We know that $|x-3|$ and $|x|$ represent distinct non-negative integers, so the sum $|x-3| + |x|$ must be larger than 1.

Thus, $|2x + 4|$ must be less than $11-1=10$ , $2 |x+2| < 10 \quad \Rightarrow \quad |x+2| < 5 \quad \Rightarrow \quad -7 < x < 3 .$

Do trial and error on these 9 possible solutions shows that $x = -3,-2,\ldots, 2$ are the only solutions. The answers is $\boxed6$ .