Step 1: Set up the vector equation.
We have $\vec{a}+3\vec{b}+4\vec{c} = x(\vec{a}-2\vec{b}+3\vec{c}) + y(\vec{a}+5\vec{b}-2\vec{c}) + z(6\vec{a}+14\vec{b}+4\vec{c})$.
Step 2: Expand the right side.
Coefficient of $\vec{a}$: $x + y + 6z$.
Coefficient of $\vec{b}$: $-2x + 5y + 14z$.
Coefficient of $\vec{c}$: $3x - 2y + 4z$.
Step 3: Equate coefficients (since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar, hence linearly independent).
From $\vec{a}$: $x + y + 6z = 1$ ... (i)
From $\vec{b}$: $-2x + 5y + 14z = 3$ ... (ii)
From $\vec{c}$: $3x - 2y + 4z = 4$ ... (iii)
Step 4: Solve equations (i) and (ii).
$2 \times$(i) + (ii): $2x+2y+12z - 2x+5y+14z = 2+3$, so $7y + 26z = 5$ ... (iv).
Step 5: Solve equations (i) and (iii).
$3 \times$(i) $-$ (iii): $3x+3y+18z - 3x+2y-4z = 3-4$, so $5y + 14z = -1$ ... (v).
From (iv) and (v): $7y+26z=5$ and $5y+14z=-1$. Multiply (v) by 7 and (iv) by 5: $35y+98z=-7$, $35y+182z=25$. Subtracting: $84z=32$, $z=\frac{32}{84}=\frac{8}{21}$. Then $y = \frac{-1-14 \cdot \frac{8}{21}}{5} = \frac{-1-\frac{16}{3}}{5} = \frac{\frac{-19}{3}}{5} = -\frac{19}{15}$. From (i): $x = 1-y-6z = 1+\frac{19}{15}-\frac{16}{7}$. Let me recompute more carefully.
Step 6: Recompute and find $x+y+z$.
Using Cramer's method or noting the answer: From the system, adding all three equations: $(1-2+3)x + (1+5-2)y + (6+14+4)z = 1+3+4$, so $2x+4y+24z = 8$, giving $x+2y+12z = 4$. From (i): $x+y+6z=1$. Subtracting: $y+6z=3$. Also from (v): $5y+14z=-1$. From $y=3-6z$: $5(3-6z)+14z=-1$, $15-30z+14z=-1$, $-16z=-16$, $z=1$. Then $y=3-6=-3$, $x=1-(-3)-6=1+3-6=-2$. So $x+y+z=-2+(-3)+1=-4$. This matches option (2).
\[ \boxed{x+y+z = -4} \]