1. For the system to have infinitely many solutions, the three equations must be consistent and dependent.

   The system is: $2x+y-z=7 \quad ...(1)$ $x-3y+2z=1 \quad ...(2)$ $x+4y+\delta z=k \quad ...(3)$

2. Write the coefficient matrix:

   2 & 1 & -1\\ 1 & -3 & 2\\ 1 & 4 & \delta \end{pmatrix}$$ For infinitely many solutions, we need: $$\det(A)=0$$ and the third equation must be a linear combination of the first two, including constants.

3. Compute the determinant: $\det(A)=2\begin{vmatrix}-3 & 2\\4 & \delta\end{vmatrix}-1\begin{vmatrix}1 & 2\\1 & \delta\end{vmatrix}+(-1)\begin{vmatrix}1 & -3\\1 & 4\end{vmatrix}$

   $=2(-3\delta-8)-(\delta-2)-\big(4-(-3)\big)$ $=2(-3\delta-8)-\delta+2-7$ $=-6\delta-16-\delta-5$ $=-7\delta-21$

   For infinitely many solutions: $-7\delta-21=0$ $\delta=-3$

4. Now substitute $\delta=-3$ into the third equation: $x+4y-3z=k$

   We now check whether the third equation is a linear combination of the first two.

   Let $a(2x+y-z)+b(x-3y+2z)=x+4y-3z$

   Comparing coefficients: $2a+b=1$ $a-3b=4$ $-a+2b=-3$

5. Solve the first two equations: From $2a+b=1 \Rightarrow b=1-2a$

   Put into $a-3(1-2a)=4$ $a-3+6a=4$ $7a=7$ $a=1$ $b=1-2=-1$

   Check the $z$-coefficient: $-a+2b=-1+2(-1)=-3$ which matches.

6. Therefore, $\text{(3)} = 1\cdot \text{(1)} - 1\cdot \text{(2)}$

   So the constant term must also satisfy: $k = 1\cdot 7 - 1\cdot 1 = 6$

7. Hence, $\delta+k=-3+6=3$

Therefore, the correct option is: $\boxed{3}$ which is Option B.