Step 1: Square the given vector sum.
Given $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. Take the dot product of each side with itself: \[ |\vec{a}+\vec{b}+\vec{c}|^2 = 0 \]
Step 2: Expand $|\vec{a}+\vec{b}+\vec{c}|^2$.
\[ |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})=0 \]
Step 3: Substitute the given magnitudes.
$|\vec{a}|=1$, $|\vec{b}|=3$, $|\vec{c}|=4$, so $|\vec{a}|^2=1$, $|\vec{b}|^2=9$, $|\vec{c}|^2=16$: \[ 1+9+16+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})=0 \] \[ 26+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})=0 \]
Step 4: Solve for the dot product sum.
\[ 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})=-26 \] \[ \vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}=-13 \]
Step 5: Interpret the result.
The negative value tells us that on average, the vectors make obtuse angles with each other, which makes sense since they must sum to zero (they form a closed triangle in vector terms).
Step 6: State the answer.
\[ \boxed{-13} \]