If the system of equations 2x + 3y - z = 5 x + + 3z = - 4 3x - y +...

Given:

The system of equations,

$2x + 3y - z = 5$

$x + \alphay + 3z = - 4$

$3x - y + \betaz = 7$ have many solutions,

So, $△ = {△}_{1} = {△}_{2} = {△}_{3}$

Now, finding {△}_{2} = \left|\begin{cases}2 & 5 & - 1 \\ 1 & - 4 & 3 \\ 3 & 7 & \beta\end{cases}\right| = 0

$⇒ 2\left(- 4\beta - 21\right) - 5\left(\beta - 9\right) - 1\left(7 + 12\right) = 0$

$⇒ 13\beta = - 16$

Now, finding {△}_{3} = \left|\begin{cases}2 & 3 & 5 \\ 1 & \alpha & - 4 \\ 3 & - 1 & 7\end{cases}\right| = 0

$⇒ 14\alpha - 8 - 57 - 5 - 15\alpha = 0$

$⇒ \alpha = - 70$

Hence, the value of $13\alpha\beta = 70 \times 16 = 1120$