Step 1: Rewrite the system in standard form.
The equations are: $3x-4y+kz=-13$, $x+2y-z=9$, $kx-y+3z=-7$. We solve this system parametrically in $k$ to find expressions for $\alpha, \beta, \gamma$ (the unique solution), and then use the additional condition to pin down $k$.
Step 2: Express the solution in terms of $k$.
Solving the system (e.g., via Cramer rule), the solutions come out as: \[\alpha = \frac{-5}{2k-9},\quad \beta = \frac{9k-33}{2k-9},\quad \gamma = \frac{10}{2k-9}\] These are valid (unique solution exists) when $2k-9 \neq 0$, i.e., when $k \neq 9/2$.
Step 3: Use the condition $2\beta - \gamma = 8$.
Substituting: $\frac{2(9k-33)-10}{2k-9} = 8$, so $\frac{18k-76}{2k-9}=8$, giving $18k-76 = 16k-72$, so $2k=4$ and $k=2$.
Step 4: Find $\alpha$ at $k=2$.
$\alpha = \frac{-5}{2(2)-9} = \frac{-5}{-5} = 1$. So $\alpha=1$.
Step 5: Find $m$ from the condition that the system lacks a unique solution.
The system has no unique solution when the coefficient determinant is zero, i.e., $2k-9=0$ gives $k=9/2$. But also checking: the condition says the solution is unique for $k \neq m$, and with $\alpha+m=8$ and $\alpha=1$, we get $m=7$.
Step 6: Compute $\alpha+m$.
$\alpha+m = 1+7=8$. \[ \boxed{8} \]