1. Let the scalar projections of $\vec b$ and $\vec c$ on $\vec a$ be equal.

   Projection condition gives $\frac{\vec a\cdot \vec b}{|\vec a|} = \frac{\vec a\cdot \vec c}{|\vec a|}$ Since $|\vec a|\neq 0$, we get $\vec a\cdot \vec b = \vec a\cdot \vec c$ Hence, $\vec a\cdot (\vec b-\vec c)=0$

2. We are also given that $\vec b \perp \vec c$, so $\vec b\cdot \vec c = 0$

3. Now compute $|\vec a+\vec b-\vec c|^2 = (\vec a+\vec b-\vec c)\cdot(\vec a+\vec b-\vec c)$

   Expanding, $|\vec a+\vec b-\vec c|^2 = |\vec a|^2+|\vec b|^2+|\vec c|^2+2\vec a\cdot\vec b-2\vec a\cdot\vec c-2\vec b\cdot\vec c$

4. Use the given conditions:

   • $|\vec a|=2 \Rightarrow |\vec a|^2=4$
   • $|\vec b|=4 \Rightarrow |\vec b|^2=16$
   • $|\vec c|=4 \Rightarrow |\vec c|^2=16$
   • $\vec a\cdot\vec b=\vec a\cdot\vec c \Rightarrow 2\vec a\cdot\vec b-2\vec a\cdot\vec c=0$
   • $\vec b\cdot\vec c=0$

   Therefore, $|\vec a+\vec b-\vec c|^2 = 4+16+16 = 36$

5. Taking square root, $|\vec a+\vec b-\vec c| = \sqrt{36}=6$

Therefore, the required integer value is $\boxed{6}$